Question 1015 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^2_{0}\text{e}^{\text{x}}\text{ dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{e}^{\text{x}},\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{0}\text{e}^{\text{x}}\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{e}^0+\text{e}^\text{h}+\text{e}^{2\text{h}}+\ .....\ +\text{e}^{(\text{n}-1)\text{h}}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\frac{(\text{e}^{\text{h}})^{\text{n}}-1}{\text{e}^{\text{h}}-1}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\text{e}^2-1}{\frac{\text{e}^{\text{h}}-1}{\text{h}}}\Bigg]$
$=\frac{\text{e}^2-1}{1}$
$=\text{e}^2-1$
View full question & answer→Question 1025 Marks
Evaluate the following definite integrals:
$\int\limits_{\text{e}}^{\text{e}^2}\Big\{\frac{1}{\log\text{x}}-\frac{1}{(\log\text{x})^2}\Big\}\text{dx}$
AnswerLet $\text{I}=\int_{\text{e}}^\limits{\text{e}^2}\Big\{\frac{1}{\log\text{x}}-\frac{1}{(\log\text{x})^2}\Big\}\text{dx}$ Then,
$\text{I}=\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{\log\text{x}}\text{ dx}-\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})^2}\text{ dx}$
Integrating by parts
$\Rightarrow\text{I}=\Bigg\{\Big[\frac{\text{x}}{\log\text{x}}\Big]^{\text{e}^2}_\text{e}-\int_{\text{e}}^\limits{\text{e}^2}\frac{-1}{\text{x}(\log\text{x})^2}\text{x dx}\Bigg\}-\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}}{\log\text{x}}\Big]^{\text{e}^2}_\text{e}+\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})^2}\text{ dx}-\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}}{\log\text{x}}\Big]^{\text{e}^2}_\text{e}+0$
$\Rightarrow\text{I}=\frac{\text{e}^2}{\log\text{e}^2}-\frac{\text{e}}{\log\text{e}}$
$\Rightarrow\text{I}=\frac{\text{e}^2}{2\log\text{e}}-\frac{\text{e}}{\log\text{e}}$
$\Rightarrow\text{I}=\frac{\text{e}^2}{2}-\text{e}$
View full question & answer→Question 1035 Marks
Evaluate the following integrals:
$\int\limits^4_1\big\{|\text{x}-1|+|\text{x}-2|+|\text{x}-4\big\}\text{dx}$
Answer$\text{I}=\int\limits^4_1\big\{|\text{x}-1|+|\text{x}-2|+|\text{x}-4\big\}\text{dx}$
$\Rightarrow\text{I}=\int\limits^4_1|\text{x}-1|\text{dx}+\int\limits^4_1|\text{x}-2|\text{dx}+\int\limits^4_1|\text{x}-4|\text{dx}$
We know that,
$|\text{x}-2|=\begin{cases}-(\text{x}-1),&\text{x}\leq1\\\text{x}-1,&1<\text{x}\leq4\end{cases}$
$|\text{x}-2|=\begin{cases}-(\text{x}-2),&1\leq\text{x}\leq2\\\text{x}-2,&2<\text{x}\leq4\end{cases}$
$|\text{x}-4|=\begin{cases}-(\text{x}-4),&1\leq\text{x}\leq4\\\text{x}-4,&\text{x}>4\end{cases}$
$\therefore\ \text{I}=\int\limits^4_1(\text{x}-1)\text{dx}-\int\limits^2_1(\text{x}-2)\text{dx}+\int\limits^4_2(\text{x}-2)\text{dx}-\int\limits^4_1(\text{x}-4)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}^2}{2}-\text{x}\Big]^4_1-\Big[\frac{\text{x}^2}{2}-\text{x}\Big]^2_1+\Big[\frac{\text{x}^2}{2}-\text{x}\Big]^4_2-\Big[\frac{\text{x}^2}{2}-\text{x}\Big]^4_1$
$\Rightarrow\text{I}=8-4-\frac{1}{2}+1-\Big(2-4-\frac{1}{2}+2\Big)+8-8-2+4-\Big(8-16-\frac{1}{2}+4\Big)$
$\Rightarrow\text{I}=\frac{23}{2}$
View full question & answer→Question 1045 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^4\text{x dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^4\text{x dx}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(\sin^2\text{x}\big)^2\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\Big(\frac{1-\cos2\text{x}}{2}\Big)^2\text{dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(1-2\cos2\text{x}+\cos^22\text{x}\big)\text{dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{x dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}(1+\cos4\text{x})\text{dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{x dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos4\text{x dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{x dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos4\text{x dx}$
$=\frac{3}{8}\Big[\text{x}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}-\frac{1}{4}\Big[\sin2\text{x}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}+\frac{1}{32}\Big[\sin4\text{x}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$
$=\frac{3}{8}\Big(\frac{\pi}{2}+\frac{\pi}{2}\Big)-\frac{1}{4}(0-0)+\frac{1}{32}(0-0)$
Hence, $\text{I}=\frac{3\pi}{8}$
View full question & answer→Question 1055 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^{2}_{0}\big(3\text{x}^2-2\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=3\text{x}^2-2,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^{2}_{0}\big(3\text{x}^2-2\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0-2)+(3\text{h}^2-2)+\ ....\ +\{3(\text{n}-1)^2\text{h}^2-2\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[-2\text{n}+3\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[-2\text{n}+3\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[-2\text{n}+\frac{2(\text{n}-1)(2\text{n}-1)}{\text{n}}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\bigg\{-2+2\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg\}$
$=-4+8$
$=4$
View full question & answer→Question 1065 Marks
Evaluate the following definite integrals:
$\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{1}{1+\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{-\frac{\pi}{4}}^\limits{\frac{\pi}{4}}\frac{1}{1+\sin\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{-\frac{\pi}{4}}^\limits{\frac{\pi}{4}}\frac{1}{1+\sin\text{x}}\times\frac{1-\sin\text{x}}{1-\sin\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{-\frac{\pi}{4}}^\limits{\pi}\frac{1-\sin\text{x}}{1-\sin^2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{-\frac{\pi}{4}}^\limits{\pi}\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{-\frac{\pi}{4}}^\limits{\pi}\Big(\frac{1}{\cos^2\text{x}}-\frac{\sin\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$\Rightarrow\text{I}=\int_{-\frac{\pi}{4}}^\limits{\pi}\big(\sec^2\text{x}-\sec\text{x}\tan\text{x}\big)\text{ dx}$
$\Rightarrow\text{I}=\big[\tan\text{x}-\sec\text{x}\big]^\frac{\pi}{4}_{-\frac{\pi}{4}}$
$\Rightarrow\text{I}=\big(1-\sqrt{2}\big)-\big(-1-\sqrt{2}\big)$
$\Rightarrow\text{I}=2$
View full question & answer→Question 1075 Marks
Evaluate the following integrals:
$\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}}{\sqrt[4]{\text{x}+4}+\sqrt[9]{9-\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}}{\sqrt[4]{\text{x}+4}-\sqrt[9]{9-\text{x}}}\text{ dx}\ ....(\text{i})$
$\text{I}=\int\limits^{5}_0\frac{\sqrt[4]{9-\text{x}}}{\sqrt[4]{9-\text{x}}-\sqrt[4]{\text{x}+4}}\text{ dx}$
$\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$\text{I}=\int\limits^{5}_0\frac{\sqrt[4]{9-\text{x}}}{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}}{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}-\frac{\sqrt[4]{9-\text{x}}}{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}\text{ dx}$
$=\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}\text{ dx}$
$=\int\limits^{5}_0\text{dx}$
$=\big[\text{x}\big]^5_0$
$=5$
Hence, $\text{I}=\frac{5}{2}$
View full question & answer→Question 1085 Marks
Evaluate the following integrals:
$\int\limits_{1}^{2}\frac{1}{\text{x}\big(1+\log\text{x}\big)^2}\text{ dx}$
AnswerLet $1+\log\text{x}=\text{t}$
Differentiating w.r.t. x, we get
$\frac{1}{\text{x}}\text{ dx}=\text{dt}$
Now, $\text{x}=1\Rightarrow\text{t}=1$
$=\text{x}=2\Rightarrow\text{t}=1+\log2$
$\therefore\ \int_{1}^\limits{2}\frac{1}{\text{x}\big(1+\log\text{x}\big)^2}\text{ dx}=\int^\limits{1+\log2}_1\frac{\text{dt}}{\text{t}^2}$
$=\Big[\frac{-1}{\text{t}}\Big]^{1+\log2}_1$
$=\bigg[\frac{-1}{1+\log2}+1\bigg]$
$=\bigg[\frac{-1+1+\log2}{1+\log2}\bigg]$
$=\bigg[\frac{\log2}{1+\log2}\bigg]$ $\big[\because\log\text{e}=1\big]$
$=\frac{\log2}{\log\text{e}+\log2}$ $\big[\log\text{a}+\log\text{b}=\log\text{ab}\big]$
$=\frac{\log2}{\log2\text{e}}$
View full question & answer→Question 1095 Marks
Evaluate the following integrals:
$\int\limits_{0}^{1}\frac{1}{1+2\text{x}+2\text{x}^2+2\text{x}^3+\text{x}^4}\text{ dx}$
Answer$\int_{0}^\limits{1}\frac{1}{1+2\text{x}+2\text{x}^2+2\text{x}^3+\text{x}^4}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{\big(\text{x}^2+1\big)^2+2\text{x}\big(\text{x}^2+1\big)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{\big(\text{x}^2+1\big)\big(\text{x}^2+1+2\text{x}\big)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{(\text{x}^2+1)(\text{x}+1)}\text{ dx}$
Let $\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{Cx}+\text{D}}{\text{x}^2+1}$
$\Rightarrow1=\text{A}(\text{x}+1)(\text{x}^2+1)+\text{B}(\text{x}^2+1)+(\text{Cx}+\text{D})(\text{x}+1)^2$
Putting x = -1, we have
$1=2\text{B}$
$\Rightarrow\text{B}=\frac{1}{2}\ ...(\text{i})$
Putting x = 0, we have
$\text{A}+\text{B}+\text{C}=1\ ...(\text{ii})$
Equating co-efficient of $x^3$ on both sides, we have
$\text{A}+\text{C}=0\ ...(\text{iii}) $
Equating co-efficient of $x^2$ on both sides, we have
$\text{A}+\text{B}+2\text{C}+\text{D}=0\ ...(\text{iv})$
$\Rightarrow2\text{C}=-1$ [Using (i)]
$\Rightarrow\text{C}=-\frac{1}{2}$
$\therefore\ \text{A}=\frac{1}{2}$ [Using (iii)]
Putting $\text{A}=\frac{1}{2},\text{ B}=\frac{1}{2}$ and $\text{C}=-\frac{1}{2}$ in (iv), we have
$\text{D}=0$
$\therefore\ \int_{0}^\limits{1}\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{\frac{1}{2}}{\text{x}+1}\text{ dx}+\int_{0}^\limits{1}\frac{\frac{1}{2}}{(\text{x}+1)^2}\text{ dx}+\int_{0}^\limits{1}\frac{-\frac{1}{2}\text{x}}{\text{x}^2+1}$
$=\Big[\frac{1}{2}\log(\text{x}+1)\Big]^1_0+\Big[\frac{1}{2}\times\Big(-\frac{1}{\text{x}+1}\Big)\Big]^1_0-\frac{1}{4}\int_{0}^\limits{1}\frac{2\text{x}}{\text{x}^2+1}\text{ dx}$
$=\frac{1}{2}\big(\log2-\log1\big)-\frac{1}{2}\Big(\frac{1}{2}-1\Big)-\Big[\frac{1}{4}\log(\text{x}^2+1)\Big]^1_0$
$=\frac{1}{2}\log2+\frac{1}{4}-\frac{1}{4}\big(\log2-\log1\big)$ $(\log1=0)$
$=\frac{1}{2}\log2+\frac{1}{4}\log\text{e}-\frac{1}{4}\log2$
$=\frac{1}{4}\log2+\frac{1}{4}\log_\text{e}$
$=\frac{1}{4}\big(\log2+\log_\text{e}\big)$
$=\frac{1}{4}\log(2\text{e})$
View full question & answer→Question 1105 Marks
Evaluate the following integrals:
$\int\limits^{8}_2\frac{\sqrt{10-\text{x}}}{\sqrt{\text{x}}+\sqrt{10-\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{8}_2\frac{\sqrt{10-\text{x}}}{\sqrt{\text{x}}+\sqrt{10-\text{x}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{8}_2\frac{\sqrt{10-(2+8-\text{x}})}{\sqrt{2+8-\text{x}}+\sqrt{10-(2+8-\text{x}})}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{8}_2\frac{\sqrt{\text{x}}}{\sqrt{10-\text{x}}+\sqrt{\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{8}_2\frac{\sqrt{10-\text{x}}}{\sqrt{\text{x}}+\sqrt{10-\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{8}_2\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^8_2$
$\Rightarrow2\text{I}=8-2=6$
$\Rightarrow\text{I}=3$
View full question & answer→Question 1115 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^{2}_{0}\big(\text{x}^2-\text{x}\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{x}^2-\text{x},\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^{2}_{0}\big(\text{x}^2-\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0-0)+(\text{h}^2-\text{h})+\ ....+\ \big\{(\text{n}-1)^2\text{h}^2-(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}\\-\text{h}\big\{1+2+\ ....+\ (\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}-\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{2}{\text{n}}\Big[\frac{2(\text{n}-1)(2\text{n}-1)}{3\text{n}}-\text{n}+1\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}2\Big[\frac{2}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-1+\frac{1}{\text{n}}\Big]$
$=\frac{8}{3}-2$
$=\frac{2}{3}$
View full question & answer→Question 1125 Marks
Evaluate the following integrals:
$\int\limits^0_{-5}\text{f(x)}\text{dx,}$ Where $\text{f(x)}=|\text{x}|+|\text{x}+2|+|\text{x}+5|$
View full question & answer→Question 1135 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^3_0(\text{x}+4)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=3,\text{ f(x)}=\text{x}+4,\text{ h}=\frac{3-0}{\text{n}}=\frac{3}{\text{n}}$
Therefore, $\text{I}=\int\limits^3_0(\text{x}+4)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}(0+(\text{n}-1)\text{h})\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0+4)+(\text{h}+4)+\ ....\ +((\text{n}-1)\text{h}+4)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}(1+2+\ ....\ +(\text{n}-1))\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{3}{\text{n}}\Big[4\text{n}+\frac{3}{\text{n}}\times\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\Big[12+\frac{9}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big]$
$=12+\frac{9}{2}$
$=\frac{33}{2}$
View full question & answer→Question 1145 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{1}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
AnswerWe have,
$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\int_{0}^\limits{\frac{\pi}{2}}\begin{pmatrix}\frac{\frac{1}{\cos^2\text{x}}}{\text{a}^2\frac{\sin^2\text{x}}{\cos^2\text{x}}+\text{b}^2\frac{\cos^2\text{x}}{\cos^2\text{x}}} \end{pmatrix}\text{dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\Big(\frac{\sec^2\text{x}}{\text{a}^2\tan^{2}\text{x}+\text{b}^2}\Big)\text{dx}$
$=\frac{1}{\text{a}^2}\int_{0}^\limits{\frac{\pi}{2}}\Bigg(\frac{\sec^2\text{x}}{\tan^{2}\text{x}+\big(\frac{\text{b}}{\text{a}}\big)^2}\Bigg)\text{dx}$
Let $\tan\text{x}=\text{t}$
Differentiating w.r.t. x, we get
$\sec^2\text{xdx}=\text{dt}$
When $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=\infty$
$\therefore\ \frac{1}{\text{a}^2}\int_{0}^\limits{\frac{\pi}{2}}\Bigg(\frac{\sec^2\text{x}}{\tan^{2}\text{x}+\big(\frac{\text{b}}{\text{a}}\big)^2}\Bigg)\text{dx}$
$=\frac{1}{\text{a}^2}\int_{0}^\limits{\infty}\frac{\text{dt}}{\big(\frac{\text{b}}{\text{a}}\big)^2+\text{t}^2}$
$=\frac{1}{\text{a}^2}\Big[\frac{\text{a}}{\text{b}}\tan^{-1}\frac{\text{at}}{\text{b}}\Big]^{\infty}_0$
$=\frac{1}{\text{a}^2}\frac{\text{a}}{\text{b}}\big[\tan^{-1}\infty-\tan^{-1}0\big]$
$=\frac{1}{\text{ab}}\Big[\tan^{-1}\tan\frac{\pi}{2}\Big]$
$=\frac{\pi}{2\text{ab}}$
View full question & answer→Question 1155 Marks
Evaluate the following integrals:
$\int\limits^{\text{b}}_{\text{a}}\frac{\text{x}^{\frac{1}{\text{n}}}}{\text{x}^\frac{1}{\text{n}}+\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}\text{ dx},\text{ n}\in\text{N},\text{n}\leq2$
AnswerLet $\text{I}=\int\limits^{\text{b}}_{\text{a}}\frac{\text{x}^{\frac{1}{\text{n}}}}{\text{x}^\frac{1}{\text{n}}+\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\text{b}}_{\text{a}}\frac{\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}{\big(\text{a}+\text{b}-\text{x}\big)^\frac{1}{\text{n}}+\big[\text{a}+\text{b}-\big(\text{a}+\text{b}-\text{x}\big)\big]^{\frac{1}{\text{n}}}}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_{\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\text{b}}_{\text{a}}\frac{\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}{\big(\text{a}+\text{b}-\text{x}\big)^\frac{1}{\text{n}}+\text{x}^{\frac{1}{\text{n}}}}\text{ dx}$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\text{b}}_{\text{a}}\frac{\text{x}^{^{\frac{1}{\text{n}}}}+\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}{\text{x}^\frac{1}{\text{n}}+\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\text{b}}_{\text{a}}\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{\text{b}}_{\text{a}}=(\text{b}-\text{a})$
$\Rightarrow\text{I}=\frac{\text{b}-\text{a}}{2}$
View full question & answer→Question 1165 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^{4}_{1}\big(\text{x}^2-\text{x}\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=1,\text{ b}=4,\text{ f(x)}=\text{x}^2-\text{x},\text{ h}=\frac{4-1}{\text{n}}=\frac{3}{\text{n}}$
Therefore, $\text{I}=\int\limits^{4}_{1}\big(\text{x}^2-\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(1)+\text{f}(1+\text{h})\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(1-1)+(1+\text{h})^2-(1+\text{h})+\\\ ....\ +\big\{(\text{n}-1)\text{h}+1\big\}^2-\big\{(\text{n}-1)\text{h}+1\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2+4\text{h}\big\}\\-\text{h}\big\{1+2+\ ....+\ (\text{n}-1)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}-\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{3}{\text{n}}\Big[\frac{3(\text{n}-1)(2\text{n}-1)}{2\text{n}}+\frac{3(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}3\Big[\frac{3}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{3}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big]$
$=9+\frac{9}{3}$
$=\frac{38}{3}$
View full question & answer→Question 1175 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\sin\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\sin\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\big(\frac{\pi}{2}-\text{x}\big)+\text{b}\cos\big(\frac{\pi}{2}-\text{x}\big)}{\sin\big(\frac{\pi}{2}-\text{x}\big)+\cos\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\cos\text{x}+\text{b}\sin\text{x}}{\cos\text{x}+\sin\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big(\frac{\text{a}\sin\text{x}+\text{b}\cos\text{x}}{\cos\text{x}+\sin\text{x}}+\frac{\text{a}\cos\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\Big)\text{dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big(\frac{\text{a}\sin\text{x}+\text{b}\cos\text{x}+\text{a}\cos\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\Big)\text{dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{(\text{a}+\text{b})\sin\text{x}+(\text{a}+\text{b})\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{(\text{a}+\text{b})(\sin\text{x}+\cos\text{x})}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0(\text{a}+\text{b})\text{dx}$
$\Rightarrow2\text{I}=(\text{a}+\text{b})\times\big[\text{x}\big]^{\frac{\pi}{2}}_0$
$\Rightarrow2\text{I}=(\text{a}+\text{b})\times\Big(\frac{\pi}{2}-0\Big)$
$\Rightarrow2\text{I}=\frac{\pi}{2}(\text{a}+\text{b})$
$\Rightarrow\text{I}=\frac{\pi}{4}(\text{a}+\text{b})$
View full question & answer→Question 1185 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^3_{1}(2\text{x}+3)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=1,\text{ b}=3,\text{ f(x)}=2\text{x}+3,\text{ h}=\frac{3-1}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^3_{1}(2\text{x}+3)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(1)+\text{f}(1+\text{h})+\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(2+3)+(2+2\text{h}+3)+\\\ ....\ +\{2+2(\text{n}-1)\text{h}+3\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[5\text{n}+2\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[5\text{n}+2\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[5\text{n}+2\text{n}-2\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big({7}-\frac{2}{\text{n}}\Big)$
$=14$
View full question & answer→Question 1195 Marks
If $\text{f}(\text{a}+\text{b}-\text{x})=\text{f(x)},$ then prove that, $\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\Big(\frac{\text{a}+\text{b}}{2}\Big)\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{x})\text{dx}$
Answer$\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})(\text{a}+\text{b}-\text{x}){}\text{dx}$ $\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\int\limits^{\text{b}}_{\text{a}}(\text{a}+\text{b}-\text{x})\text{f}(\text{x})\text{dx}$ $\big[\text{f}(\text{a}+\text{b}-\text{x})=\text{f}(\text{x})\big]$
$\Rightarrow\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\int\limits^{\text{b}}_{\text{a}}(\text{a}+\text{b})\text{f}(\text{x})\text{dx}-\int\limits^{\text{b}}_{\text{a}}\text{x}\text{f}(\text{x})\text{dx}$
$\Rightarrow2\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=(\text{a}+\text{b})\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{x})\text{dx}$
$\Rightarrow\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\Big(\frac{\text{a}+\text{b}}{2}\Big)\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{x})\text{dx}$
View full question & answer→Question 1205 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{4}}_{0}\frac{\sin^2\text{x}\cos^2\text{x}}{(\sin^3\text{x}\cos^3\text{x})}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{4}}_{0}\frac{\sin^2\text{x}\cos^2\text{x}}{(\sin^3\text{x}\cos^3\text{x})}\text{ dx}$
$=\int^\limits{\frac{\pi}{4}}_{0}\frac{\sin^2\text{x}\cos^2\text{x}}{\cos^6\text{x}(\tan^3\text{x}+1)^2}\text{ dx}=\int^\limits{\frac{\pi}{4}}_{0}\frac{\tan^2\text{x}\sec^2\text{x}}{(\tan^3\text{x}+1)}\text{ dx}$
Put $\tan^3\text{x}+1=\text{z}$
$\therefore\ 3\tan^{2}\text{x}\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\tan^{2}\text{x}\sec^2\text{x dx}=\frac{\text{dz}}{3}$
When $\text{x}\rightarrow0,\text{z}\rightarrow1$
When $\text{x}\rightarrow\frac{\pi}{4},\text{z}\rightarrow2$
$\therefore\ \text{I}=\frac{1}{3}\int^\limits{2}_1\frac{\text{dz}}{\text{z}^2}$
$=\frac{1}{3}\times-\Big[\frac{1}{\text{z}}\Big]^2_1$
$=-\frac{1}{3}\Big(\frac{1}{2}-1\Big)$
$=-\frac{1}{3}\times\Big(-\frac{1}{2}\Big)$
$=\frac{1}{6}$
View full question & answer→Question 1215 Marks
Evaluate the following integrals:
$\int\limits^2_{0}\big|\text{x}^2-3\text{x}+2\big|\text{dx}$
Answer$\int^\limits2_{0}\big|\text{x}^2-3\text{x}+2\big|\text{dx}$
We know that,
$\big|\text{x}^2-3\text{x}+2\big|\text{dx}=\begin{cases}-(\text{x}^2-3\text{x}+2),&(\text{x}-1)(\text{x}-2)\leq0\text{ or },&1\leq\text{x}\leq2\$\text{x}^2-3\text{x}+2),&\text{x}^2-3\text{x}+2\leq0\text{ or },&\text{x}\in(-\infty,1)(2,\infty)\end{cases}$
$\therefore\ \text{I}=\ \int^\limits2_{0}\big(\text{x}^2-3\text{x}+2\big)\text{dx}$
$\Rightarrow\text{I}=\int^\limits1_{0}\big(\text{x}^2-3\text{x}+2\big)\text{dx}-\int^\limits2_1\big(\text{x}^2-3\text{x}+2\big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}^3}{3}-\frac{3\text{x}^2}{2}+2\text{x}\Big]^1_0-\Big[\frac{\text{x}^3}{3}-\frac{3\text{x}^2}{2}+2\text{x}\Big]^2_1$
$\Rightarrow\text{I}=\frac{1}{3}-\frac{3}{2}+2-\Big[\frac{8}{3}-6+4-\frac{1}{3}+\frac{3}{2}-2\Big]$
$\Rightarrow\text{I}=\frac{1}{3}-\frac{3}{2}+2-\frac{8}{3}+6-2+\frac{1}{3}-\frac{3}{2}$
$\Rightarrow\text{I}=1$
View full question & answer→Question 1225 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\sin2\text{x }\tan^{-1}(\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_0\sin2\text{x }\tan^{-1}(\sin\text{x})\text{dx}$
Differentiating w.r.t. x, we get
$\cos\text{x dx}=\text{dt}$
Now, $\text{x}=0,\text{t}=0$
$\text{x}=\frac{\pi}{2},\text{t}=1$
$\therefore\ \int^\limits{\frac{\pi}{2}}_0\sin2\text{x }\tan^{-1}(\sin\text{x})\text{dx}=2\int^\limits{1}_0\text{t }\tan^{-1}\text{t dt}$ $\big[\because\sin2\text{x}=2\sin\text{x }\cos\text{x}\big]$
Using by parts
$=2\Big\{\tan^{-1}\text{t}\int\text{t dt}-\int\big(\int\text{t dt}\big)\frac{\text{d}\tan^{-1}\text{t}}{\text{dt}}\Big\}$
$=2\Big\{\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\frac{1}{2}\int\frac{\text{t}^2}{1+\text{t}^2}\text{ dt}\Big\}$
$=2\bigg\{\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\frac{1}{2}\Big(\int\text{dt}-\int\frac{\text{dt}}{1+\text{t}^2}\text{ dt}\Big)\bigg\}$
$=2\Big[\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\frac{1}{2}\big(\text{t}-\tan^{-1}\text{t}\big)\Big]^1_0$
$=2\bigg\{\frac{1}{2}\frac{\pi}{4}-\frac{1}{2}\Big(1-\frac{\pi}{4}\Big)\bigg\}$
$=2\Big\{\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}\Big\}$
$=2\Big(\frac{\pi}{4}-\frac{1}{2}\Big)$
$=\frac{\pi}{2}-1$
View full question & answer→Question 1235 Marks
Evaluate the following integrals:
$\int\limits^{\frac{3}{2}}_0\big|\text{x}\cos\pi\text{x}\big|\text{dx}$
Answer$\int\limits^\frac{3}{2}_{0}|\text{x} \cos\pi \text{ x}| \text{dx} = \int\limits^{1/2}_{0}\text{x}\cos\pi \text{x dx}{-}\int\limits^{3/2}_{1/2}\text{x}\cos\pi \text{x dx}$
$= \left\{\frac{\text{x}\sin\pi \text{x}}{\pi} + \frac{\cos\pi\text{x}}{\pi^{2}}\right\}^{1/2}_{0}= \left\{\frac{\text{x}\sin\pi \text{x}}{\pi} + \frac{\cos\pi\text{x}}{\pi^{2}}\right\}^{3/2}_{1/2}$
$=\frac{1}{2\pi} - \frac{1}{\pi^{2}} - \bigg(-\frac{3}{2\pi}-\frac{1}{2\pi}\bigg) = \frac{5}{2\pi}-\frac{1}{\pi^{2}}$
View full question & answer→Question 1245 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\text{n}}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx},\text{ n}\in\text{N}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\text{n}}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\text{n}}\big(\frac{\pi}{2}-\text{x}\big)}{\sin^{\text{n}}\big(\frac{\pi}{2}-\text{x}\big)+\cos^{\text{n}}\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\cos^\text{n}\text{x}+\sin^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\bigg[\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}+\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\bigg]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\big[\text{x}\big]^{\frac{\pi}{2}}_0$
$=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
View full question & answer→Question 1255 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\cos^5\text{x dx}$
Answer$\int^\limits{\frac{\pi}{2}}_0\cos^5\text{x dx}=\int^\limits{\frac{\pi}{2}}_0\big(1-\sin^2\text{x}\big)^2\cos\text{x dx}$
Let $\sin\text{x}=\text{t}$
Differentiating w.r.t. x, we get
$\cos\text{x dx}=\text{dt}$
When $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=1$
$=\int^\limits{\frac{\pi}{2}}_0\big(1-\sin^2\text{x}\big)^2\cos\text{x dx}$
$=\int^\limits{1}_0\big(1-\text{t}^2\big)^2\text{ dt}$
$=\int^\limits{1}_0\big(1-2\text{t}^2+\text{t}^4\big)\text{dt}$
$=\Big[\text{t}-\frac{2}{3}\text{t}^3+\frac{\text{t}^5}{5}\Big]^1_0$
$=1-\frac{2}{3}+\frac{1}{5}$
$=\frac{8}{15}$
View full question & answer→Question 1265 Marks
Evaluate the following definite integrals:
$\int\limits_{\frac{\pi}{2}}^{\pi}\text{e}^{\text{x}}\Big(\frac{1-\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{\frac{\pi}{2}}^\limits{\pi}\text{e}^{\text{x}}\Big(\frac{1-\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}$ Then,
$\text{I}=\int_{\frac{\pi}{2}}^\limits{\pi}\text{e}^{\text{x}}\bigg(\frac{1-2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\sin^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int_{\frac{\pi}{2}}^\limits{\pi}\text{e}^{\text{x}}\Big(\frac{1}{2}\text{cosec}^2\frac{\text{x}}{2}-\cot\frac{\text{x}}{2}\Big)\text{dx}$
$\Rightarrow\text{I}=\int_{\frac{\pi}{2}}^\limits{\pi}\frac{1}{2}\text{e}^{\text{x}}\text{cosec}^2\frac{\text{x}}{2}\text{ dx}-\int_{\frac{\pi}{2}}^\limits{\pi}\text{e}^{\text{x}}\cot\frac{\text{x}}{2}\text{ dx}$
Integrating second term by parts,
$\text{I}=\bigg\{-\Big[\text{e}^{\text{x}}\cot\frac{\text{x}}{2}\Big]^{\pi}_\frac{\pi}{2}-\int_{\frac{\pi}{2}}^\limits{\pi}\frac{1}{2}\text{ e}^{\text{x}}\text{cosec}^2\frac{\text{x}}{2}\text{ dx}\bigg\}+\int_{\frac{\pi}{2}}^\limits{\pi}\frac{1}{2}\text{ e}^{\text{x}}\text{cosec}^2\frac{\text{x}}{2}\text{ dx}$
$\Rightarrow\text{I}=-\Big[0-\text{e}^{\frac{\pi}{2}}\Big]$
$\Rightarrow\text{I}=\text{e}^{\frac{\pi}{2}}$
View full question & answer→Question 1275 Marks
Evaluate the following integrals:
$\int\limits^{\frac{3}{2}}_0\big|\text{x}\sin\pi\text{x}\big|\text{dx}$
AnswerFor $0<\text{x}<1,\text{ x}>0$ and $\sin\pi\text{x}>0\Rightarrow\text{x}\sin\pi\text{x}>0$
For $1<\text{x}<\frac{3}{2},\text{ x}>0$ and $\sin\pi\text{x}>0\Rightarrow\text{x}\sin\pi\text{x}>0$
$\therefore\ \int\limits^{\frac{3}{2}}_0\big|\text{x}\sin\pi\text{x}\big|\text{dx}=\int\limits^1_0\text{x}\sin\pi\text{x dx}-\int\limits^{\frac{3}{2}}_1\text{x}\sin\pi\text{x dx}$
Let $\text{I}=\int\text{x}\sin\pi\text{ dx}$
$=\text{x}\int\sin\pi\text{x}-\int\Big(\frac{\text{d}}{\text{dx}}\text{x}\int\sin\pi\text{x dx}\Big)\text{dx}$
$=\text{x}\Big(\frac{-\cos\pi\text{x}}{\pi}\Big)-\int\Big(\frac{-\cos\pi\text{x}}{\pi}\Big)\text{dx}$
$=\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}$
Applying the limits, we get
$\int\limits^{\frac{3}{2}}_0|\text{x}\sin\pi\text{x}|\text{dx}=\Big[\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}\Big]^1_0-\Big[\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}\Big]^{\frac{3}{2}}_1$
$=\Big[\Big(\frac{\cos\text{x}}{\pi}+\frac{\sin\pi}{\pi^2}\Big)-(0+0)\Big]-\Bigg[\bigg(\frac{-\frac{3}{2}\cos\frac{3\pi}{2}}{\pi}+\frac{\sin\frac{3\pi}{2}}{\pi^2}\bigg)-\Big(\frac{-\cos\pi}{\pi}+\frac{\sin\pi}{\pi^2}\Big)\Bigg]$
$=\bigg[\Big(\frac{1}{\pi}+0\Big)\bigg]-\bigg[\Big(0-\frac{1}{\pi^2}\Big)-\Big(\frac{1}{\pi}+0\Big)\bigg]$
$=\frac{1}{\pi}+\frac{1}{\pi^2}+\frac{1}{\pi}$
$=\frac{2\pi+1}{\pi^2}$
View full question & answer→Question 1285 Marks
Evaluate the following integrals:
$\int\limits^4_{0}\big(|\text{x}|+|\text{x}+2|+|\text{x}+4|\big)\text{dx}$
Answer$\text{I}=\int\limits^4_{0}\big\{|\text{x}|+|\text{x}+2|+|\text{x}+4|\big\}\text{dx}$
$\Rightarrow\text{I}=\int\limits^4_0|\text{x}|\text{dx}+\int\limits^4_0|\text{x}-2|\text{dx}+\int\limits^4_0|\text{x}-4|\text{dx}$
We know that,
$|\text{x}|=\begin{cases}-\text{x},&-5\leq\text{x}\leq0\\\text{x},&\text{x}>0\end{cases}$
$|\text{x}-2|=\begin{cases}-(\text{x}-2),&0\leq\text{x}\leq2\\\text{x}-2,&2<\text{x}\leq4\end{cases}$
$|\text{x}-4|=\begin{cases}-(\text{x}-4),&0\leq\text{x}\leq4\\\text{x}-4,&\text{x}>4\end{cases}$
$\therefore\ \text{I}=\int\limits^4_{0}\text{x dx}-\int\limits^2_{0}(\text{x}-2)\text{dx}+\int\limits^4_{2}(\text{x}-2)\text{dx}-\int\limits^4_{0}(\text{x}-4)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}^2}{2}\Big]^4_0-\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^2_0+\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^4_2-\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^4_0$
$\Rightarrow\text{I}=8-(2-4)+8-8-2+4-(8-16)$
$\Rightarrow\text{I}=20$
View full question & answer→Question 1295 Marks
Evaluate the following integrals:
$\int\limits^1_0\log\Big(\frac{1}{\text{x}}-1\Big)\text{dx}$
AnswerLet $\text{I}=\int\limits^1_0\log\Big(\frac{1}{\text{x}}-1\Big)\text{dx}\ ....(\text{i})$
$=\int\limits^1_0\log\Big(\frac{1}{1-\text{x}}-1\Big)\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$\text{I}=\int\limits^1_0\log\Big(\frac{\text{x}}{\text{x}-1}\Big)\text{dx}\ ....(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^1_0\log\Big(\frac{1-\text{x}}{\text{x}}\Big)\log\Big(\frac{1-\text{x}}{\text{x}}\Big)\text{dx}$
$=\int\limits^1_0\log\Big(\frac{1-\text{x}}{\text{x}}\times\frac{\text{x}}{1-\text{x}}\Big)\text{dx}$
$=\int\limits^1_0\log1\text{ dx}$
$=0$
Hence, $\text{I}=0$
View full question & answer→Question 1305 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cot\text{x}}}{\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cot\text{x}}}{\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cot\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\cot\big(\frac{\pi}{2}-\text{x}\big)}+\sqrt{\tan\big(\frac{\pi}{2}-\text{x}\big)}}\text{ dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan\text{x}}+\sqrt{\cot\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\bigg(\frac{\sqrt{\cot\text{x}}}{\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}+\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan\text{x}}+\sqrt{\cot\text{x}}}\bigg)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
View full question & answer→Question 1315 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{6}}_{0}\cos^{-3}2\theta\sin2\theta\text{ d}\theta$
AnswerWe have,
$\int^\limits{\frac{\pi}{6}}_{0}\cos^{-3}2\theta\sin2\theta\text{ d}\theta$
$=\int^\limits{\frac{\pi}{6}}_{0}\frac{\sin2\theta}{\cos^32\theta}\text{ d}\theta$
$=\int^\limits{\frac{\pi}{6}}_{0}\tan2\theta\cdot\sec^22\theta\text{ d}\theta$
Let $\tan2\theta=\text{t}$
Differentiating w.r.t. x, we get
$2\sec^22\theta\text{d}\theta=\text{dt}$
Now, $\theta=0\Rightarrow\text{t}=0$
$\theta=\frac{\pi}{6}\Rightarrow\text{t}=\sqrt{3}$
$\therefore\ \int^\limits{\frac{\pi}{6}}_{0}\tan2\theta\cdot\sec^22\theta\text{ d}\theta=\frac{1}{2}\int^\limits{\sqrt{3}}_0\text{t dt}=\frac{1}{2}\Big[\frac{\text{t}^2}{2}\Big]^{\sqrt{\text{3}}}_0$
$=\frac{3}{4}$
View full question & answer→Question 1325 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^1_{-1}(\text{x}+3)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=-1,\text{ b}=1,\text{ f(x)}=\text{x}+3,\text{ h}=\frac{1+1}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^1_{-1}(\text{x}+3)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(-1)+\text{f}(-1+\text{h})+\\\ ....\ +\text{f}\big\{-1+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(-1+3)+(-1+\text{h}+3)+\\ ....\ +\{-1+(\text{n}-1)\text{h}+3\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\text{n}+\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[\text{n}+\text{n}-1\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big(3-\frac{1}{\text{n}}\Big)$
$=6$
View full question & answer→Question 1335 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^{3}_{1}\big(2\text{x}^2+5\text{x}\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=1,\text{ b}=3,\text{ f(x)}=2\text{x}^2+5\text{x},\text{ h}=\frac{3-1}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^{3}_{1}\big(2\text{x}^2+5\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(1)+\text{f}(1+\text{h})\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(2+5)+\big\{2(1+\text{h})^2+5(1+\text{h})\big\}+\ \\....+\ \big\{2(1+(\text{n}-1)\text{h}^2+5(1+(\text{n}-1)\text{h})\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\Big\{1^2+(1+\text{h}^2)+\ ....+\ \big\{1+(\text{n}-1)\text{h}\big\}^2\Big\}+\\5\big\{1+(1+\text{h})+(1+2\text{h}+\ ....+\ (1+(\text{n}+1)\text{h}))\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\text{n}+2\text{h}^2(1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2+\\4\text{h}\big\{1+2+\ ....+ (\text{n}-1)\big\}+5\text{n}+5\text{h}\big\{1+2+\ ...+\ (\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[7\text{n}+2\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+9\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{2}{\text{n}}\Big[7\text{n}+\frac{4(\text{n}-1)(2\text{n}-1)}{3\text{n}}+9\text{n}-9\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}2\Big[16+\frac{4}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{9}{\text{n}}\Big]$
$=32+\frac{16}{3}$
$=\frac{112}{3}$
View full question & answer→Question 1345 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\text{x}\cos^2\text{x dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\text{x}\cos^2\text{x dx}\ ...(\text{i})$
$=\int\limits^{\pi}_0(\pi-\text{x})\cos^2(\pi-\text{x})\text{dx}$
$=\int\limits^{\pi}_0(\pi-\text{x})\cos^2\text{x}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\cos^2\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\cos^2\text{x}\text{ dx}$
$=\pi\int\limits^{\pi}_0\frac{1+\cos2\text{x}}{2}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\big(1+\cos2\text{x}\big)\text{dx}$
$=\frac{\pi}{2}\Big[\text{x}+\frac{\sin2\text{x}}{2}\Big]^{\pi}_0$
$=\frac{\pi}{2}(\pi-0)$
Hence, $\text{I}=\frac{\pi^2}{4}$
View full question & answer→Question 1355 Marks
Evaluate the following integrals:
$\int\limits^2_{-2}|2\text{x}+3|\text{dx}$
Answer$\int^\limits2_{-2}|2\text{x}+3|\text{dx}$
We know that,
$|2\text{x}+3|=\begin{cases}-(2\text{x}+3),&-2\leq\text{x}\leq-\frac{3}{2}\$2\text{x}+3),&-\frac{3}{2}<\text{x}\leq2\end{cases}$
$\therefore\ \text{I}=\int^\limits{\frac{-3}{2}}_{-2}-\big(2\text{x}+3\big)\text{dx}+\int^\limits2_{-\frac{3}{2}}\big(2\text{x}+3\big)\text{dx}$
$\Rightarrow\text{I}=-\Big[\text{x}^3+3\text{x}\Big]^{\frac{-3}{2}}_{-2}+\Big[\text{x}^2+3\text{x}\Big]^2_{-\frac{3}{2}}$
$\Rightarrow\text{I}=-\frac{9}{4}+\frac{9}{2}+4-6+4+6-\frac{9}{4}+\frac{9}{2}$
$\Rightarrow\text{I}=\frac{25}{2}$
View full question & answer→Question 1365 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sin^2\text{x dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sin^2\text{x dx}$
Here $\text{f(x)}=\sin^2\text{x}$
$\text{f}(-\text{x})=\sin^2(-\text{x})=\sin^2\text{x}=\text{f(x)}$
Hence $\sin^2\text{x}$ is an even function
Therefore,
$\text{I}=2\int\limits^{\frac{\pi}{4}}_{0}\sin^2\text{x dx}$
$=2\int\limits^{\frac{\pi}{4}}_{0}\Big(\frac{1-\cos2\text{x}}{2}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{4}}_{0}(1-\cos2\text{x})\text{dx}$
$=\Big[\text{x}-\frac{\sin^2\text{x}}{2}\Big]^{\frac{\pi}{4}}_0$
$=\frac{\pi}{4}-\frac{1}{2}$
View full question & answer→Question 1375 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}$
AnswerWe have,
$\int_{0}^\limits{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}$ $\big[\because2\cos\text{C}\cos\text{D}=\cos(\text{C}+\text{D})-\cos(\text{C}-\text{D})\big]$
$=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{6}}2\cos\text{x }\cos2\text{x dx}$
$=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{6}}(\cos3\text{x}+\cos\text{x})\text{dx}$
$=\frac{1}{2}\int\Big[\frac{\sin3\text{x}}{3}+\sin\text{x}\Big]_0^{\frac{\pi}{6}}$
$=\frac{1}{2}\Bigg[\bigg(\frac{\sin3\frac{\pi}{6}}{3}+\sin\frac{\pi}{6}\bigg)-(\sin0-\sin0)\Bigg]$
$=\frac{1}{2}\bigg[\frac{\sin\frac{\pi}{2}}{3}+\sin\frac{\pi}{6}\bigg]$
$=\frac{1}{2}\Big(\frac{1}{3}+\frac{1}{2}\Big)$
$=\frac{1}{2}\Big(\frac{5}{6}\Big)$
$=\frac{5}{12}$
$\therefore\ \int_{0}^\limits{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}=\frac{5}{12}$
View full question & answer→Question 1385 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{0}\frac{\cos^2\text{x}}{1+3\sin^3\text{x}}\text{ dx}$
Answer$\text{I}=\int^\limits\frac{\pi}{2}_{0}\frac{\cos^2\text{x}}{1+3\sin^3\text{x}}\text{ dx}$
$\text{I}=\int^\limits\frac{\pi}{2}_{0}\frac{\sec^2\text{x}}{\sec^2\text{x}\big(\sec^2\text{x}+3\tan^2\text{x}\big)}\text{ dx}$
Put $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{x}=0\Rightarrow\text{t}=0$ and $\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=\infty$
$\Rightarrow\text{I}=\int^{\infty}\limits_0\frac{1}{\big(1+\text{t}^2\big)\big(1+4\text{t}^2\big)}\text{ dt}$
$\Rightarrow\text{I}=-\frac{1}{3}\int^{\infty}\limits_0\bigg[\frac{1}{(1+\text{t}^2)-{(1+4\text{t}^2)}}\bigg]\text{dt}$
$\Rightarrow\text{I}=-\frac{1}{3}\Big[\tan^{-1}\text{t}-2\tan^{-1}2\text{t}\Big]^{\infty}_0$
$\Rightarrow\text{I}=\frac{\pi}{6}$
View full question & answer→Question 1395 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\cot^{\frac{3}{2}}\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\cot^{\frac{3}{2}}\text{x}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\cot^{\frac{3}{2}}\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\cot^{\frac{3}{2}}\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\tan^{\frac{3}{2}}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{\cot^{\frac{3}{2}}\text{x}}{\cot^{\frac{3}{2}}\text{x}+1}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1+\cot^{\frac{3}{2}}\text{x}}{1+\cot^{\frac{3}{2}}\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{\frac{\pi}{3}}_\frac{\pi}{6}$
$\Rightarrow2\text{I}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$
$\Rightarrow\text{I}=\frac{\pi}{12}$
View full question & answer→Question 1405 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\sin^3\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\pi}_0\sin^3\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$ Then,
$\text{I}=\int^\limits{\pi}_0\sin\text{x }\sin^2\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
$\Rightarrow\text{I}=\int^\limits{\pi}_0\sin\text{x}(1-2\cos^2\text{x})(1+\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
$\Rightarrow\text{I}=\int^\limits{\pi}_0\sin\text{x}(1-\cos\text{x})(1+2\cos\text{x})(1+\cos\text{x})^3\text{ dx}$
Let $\cos\text{x}=\text{t}$ Then, $-\sin\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=1$ and $\text{x}=\pi,\text{t}=-1$
$\therefore\ \text{I}=-\int^\limits{-1}_1(1-\text{t})(1+2\text{t})(1+\text{t})^3\text{ dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+\text{t}-2\text{t}^2\big)\big(1+\text{t}^3+3\text{t}+3\text{t}^2\big)\text{dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+\text{t}^3+3\text{t}+3\text{t}^3+\text{t}+\text{t}^4\\+3\text{t}^2+3\text{t}^3-2\text{t}^2-2\text{t}^5-6\text{t}^3-6\text{t}^4\big)\text{dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+4\text{t}+4\text{t}^2-2\text{t}^3-5\text{t}^4-2\text{t}^5\big)\text{dt}$
$\Rightarrow\text{I}=\Big[\text{t}+2\text{t}^2+\frac{4\text{t}^3}{3}-\frac{\text{t}^4}{2}-\text{t}^5-\frac{\text{t}^6}{3}\Big]^1_{-1}$
$\Rightarrow\text{I}=1+2+\frac{4}{3}-\frac{1}{2}-1-\frac{1}{3}\\+1-2+\frac{4}{3}+\frac{1}{2}-1+\frac{1}{3}$
$\Rightarrow\text{I}=\frac{8}{3}$
View full question & answer→Question 1415 Marks
Evaluate the following definite integrals:
$\int\limits_{1}^{2}\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{2}\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{1}^\limits{2}\Big(\frac{\text{e}^{\text{x}}}{\text{x}}-\frac{\text{e}^{\text{x}}}{\text{x}^2}\Big)\text{dx}$
$\Rightarrow\text{I}=\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}}\text{ dx}-\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}^2}\text{ dx}$
Integrating first term by parts,
$\text{I}=\bigg\{\Big[\frac{\text{e}^\text{x}}{\text{x}}\Big]^2_1-\int_{1}^\limits{2}\frac{-1}{\text{x}^2}\text{e}^{\text{x}}\text{ dx}\bigg\}-\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{e}^\text{x}}{\text{x}}\Big]^2_1+\int_{1}^\limits{2}\frac{-1}{\text{x}^2}\text{e}^{\text{x}}\text{ dx}-\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{e}^\text{x}}{\text{x}}\Big]^2_1$
$\Rightarrow\text{I}=\frac{\text{e}^2}{2}-\text{e}$
View full question & answer→Question 1425 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan\text{x}}+\sqrt{\cot\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan\text{x}}+\sqrt{\cot\text{x}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}{\sqrt{\sin\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}+\sqrt{\cos\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_{\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}+\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\text{dx}$
$\Rightarrow2\text{I}=\Big[\text{x}\Big]^{\frac{\pi}{3}}_{\frac{\pi}{6}}$
$\Rightarrow2\text{I}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$
$\Rightarrow\text{I}=\frac{\pi}{12}$
View full question & answer→Question 1435 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{4}}\frac{\tan^{3}\text{x}}{1+\cos2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\tan^{3}\text{x}}{1+\cos2\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\tan^{3}\text{x}}{1+\cos2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{4}}\tan^{3}\text{x}\sec^2\text{x}\text{ dx}$
Let $\tan\text{x}=\text{t}$ Then, $\sec^2\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{4},\text{t}=1$
$\therefore\ \text{I}=\frac{1}{2}\int_{0}^\limits{1}\text{t}^3\text{ d}t$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\frac{\text{t}^4}{4}\Big]^1_0$
$\Rightarrow\text{I}=\frac{1}{2}\Big(\frac{1}{4}-0\Big)$
$\Rightarrow\text{I}=\frac{1}{8}$
View full question & answer→Question 1445 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\text{x}\sin^3\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\text{x}\sin^3\text{x}\text{ dx}\ ...(\text{i})$$=\int\limits^{\pi}_0(\pi-\text{x})\sin^3(\pi-\text{x})\text{dx}$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin^3\text{x}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\sin^3\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\sin^3\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\frac{3\sin\text{x}-\sin3\text{x}}{4}\text{ dx}$
$=\frac{\pi}{4}\int\limits^{\pi}_0\big(3\sin\text{x}-\sin3\text{x}\big)\text{dx}$
$=\frac{\pi}{4}\Big[-3\cos\text{x}+\frac{\cos3\text{x}}{3}\Big]^{\pi}_0$
$=\frac{\pi}{4}\Big[-3\cos\pi+3\cos0+\frac{\cos3\pi}{3}-\frac{\cos0}{3}\Big]$
$=\frac{\pi}{4}\Big[3+3+\frac{-1}{3}-\frac{1}{3}\Big]$
$=\frac{\pi}{4}\Big[3-\frac{1}{3}\Big]$
$=\frac{\pi}{2}\times\frac{8}{3}$
$=\frac{4\pi}{3}$
$\therefore\ \text{I}=\frac{2\pi}{3}$
View full question & answer→Question 1455 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\text{a}}\frac{\text{x}}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}$
AnswerLet $\text{a}^2+\text{x}^2=\text{t}^2$
Differentiating w.r.t. x, we get
$2\text{xdx}=2\text{tdt}$
$\text{xdx}=\text{tdt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\text{a}\Rightarrow\text{t}=\sqrt{2}\text{a}$
$\therefore\ \int_{0}^\limits{\text{a}}\frac{\text{x}}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}=\int_{\text{a}}^\limits{\sqrt{2\text{a}}}\frac{\text{t dt}}{\text{t}}$
$=\int_{\text{a}}^\limits{\sqrt{2\text{a}}}\text{dt}$
$=\big[\text{t}\big]^{\sqrt{2}\text{a}}_\text{a}$
$=\big[\sqrt{2}\text{a}-\text{a}\big]$
$=\text{a}\big(\sqrt{2}-1\big)$
View full question & answer→Question 1465 Marks
Evaluate the following integrals:
$\int\limits^{\frac{1}{2}}_{0}\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$
Answer$\text{I}=\int^\limits{\frac{1}{2}}_{0}\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$
Let $\text{x}=\sin\text{u}$
$\text{dx}=\cos\text{u du}$
$\text{I}=\int^\limits{\frac{\pi}{6}}_{0}\frac{1}{(1+\sin^2\text{u})}\text{ du}$
$\text{I}=\int^\limits{\frac{\pi}{6}}_{0}\frac{\sec^2\text{u}}{(1+2\tan^2\text{u})}\text{ du}$
Let $\tan\text{u}=\text{v}$
$\text{dv}=\sec^2\text{u du}$
$\text{I}=\int^\limits{\frac{1}{\sqrt{3}}}_{0}\frac{1}{(1+2\text{v}^2)}\text{ dv}$
$\text{I}=\frac{1}{\sqrt{2}}\Big[\tan^{-1}\big(\sqrt{2}\text{v}\big)\Big]^{\frac{1}{\sqrt{3}}}_0$
$\text{I}=\frac{1}{\sqrt{2}}\Big[\tan^{-1}\Big(\sqrt{\frac{2}{3}}\Big)\Big]$
View full question & answer→Question 1475 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{1}\Big(\text{xe}^{\text{x}}+\cos\frac{\pi\text{x}}{4}\Big)\text{dx}$
AnswerWe have,
$\int_{0}^\limits{1}\Big(\text{xe}^{\text{x}}+\cos\frac{\pi\text{x}}{4}\Big)\text{dx}$
$=\int_{0}^\limits{1}\text{xe}^{\text{x}}\text{ dx}+\int_{0}^\limits{1}\cos\frac{\pi\text{x}}{4}\text{ dx}$
Applying by parts in $1^{st}$ integral we get,
$=\text{x}\int_{0}^\limits{1}\text{e}^{\text{x}}\text{ dx}-\int_{0}^\limits{1}\big(\int\text{e}^{\text{x}}\text{ dx}\big)+\int_{0}^\limits{1}\cos\frac{\pi\text{x}}{4}\text{ dx}$
$=\big[\text{xe}^{\text{x}}\big]^1_0-\int_{0}^\limits{1}\text{e}^{\text{x}}\text{ dx}+\bigg[\frac{\sin\frac{\pi\text{x}}{4}}{\frac{\pi}{4}}\bigg]^1_0$
$=\big[\text{xe}^{\text{x}}-\text{e}^{\text{x}}\big]^1_0+\frac{4}{\pi}\Big[\frac{1}{\sqrt{2}}-0\Big]$
$=\big[\text{e}^{\text{x}}(\text{x}-1)\big]^1_0+\frac{4}{\pi}\Big[\frac{1}{\sqrt{2}}\Big]$
$=0+1+\frac{4}{\pi\sqrt{2}}$
$=1+\frac{2\sqrt{2}}{\pi}$
View full question & answer→Question 1485 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\text{x}^2\cos^2\text{x}\text{ dx}$
AnswerWe have,
$\int\text{x}^2\cos^2\text{x dx}=\int\text{x}^2\Big(\frac{1+\cos2\text{x}}{2}\Big)\text{dx}\\=\frac{1}{2}\int(\text{x}^2+\text{x}^2\cos2\text{x})\text{dx}=\frac{1}{2}\big[\int\text{x}^2\text{dx}+\int\text{x}^2\cos2\text{x dx}\big]\ ...(\text{A})$
Now,
$\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\text{ dx}=\Big[\frac{\text{x}^3}{3}\Big]^{\frac{\pi}{2}}_0=\frac{\pi^3}{24}\ ....(\text{B})$
$\int\text{x}^2\cos2\text{x dx}=\text{x}^2\int\cos2\text{x dx}-\int2\text{x}\big(\int\cos2\text{x dx}\big)$
$=\frac{\text{x}^2\sin2\text{x}}{2}-\int\frac{\sin2\text{x}}{2}2\text{x dx}$
$=\frac{\text{x}^2\sin2\text{x}}{2}-\Big[\text{x}\int\sin2\text{x}-\int\big(\int\sin2\text{x dx}\big)\text{dx}\Big]$
$=\frac{\text{x}^2\sin2\text{x}}{2}+\frac{\text{x}\cos2\text{x}}{2}-\int\frac{\cos2\text{x}}{2}\text{ dx}$
$\therefore\ \int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos2\text{x dx}=\Big[\frac{\text{x}^2\sin2\text{x}}{2}+\frac{\text{x}\cos2\text{x}}{2}-\frac{\sin2\text{x}}{4}\Big]^{\frac{\pi}{2}}_0=\frac{-\pi}{4}\ ...(\text{C})$
Now, put (B) & (C) in (A), we get
$\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos^2\text{x}\text{ dx}=\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\text{ dx}+\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos2\text{x dx}\\=\frac{1}{2}\Big[\frac{\pi^2}{24}-\frac{\pi}{4}\Big]=\frac{\pi^3}{48}-\frac{\pi}{8}$
View full question & answer→Question 1495 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{\pi}\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}$
Integrating by parts, we get
$\text{I}=\frac{1}{2}\Big[\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^\pi_0-\frac{1}{2}\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\cos\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}$
Now, integrating the second term by parts, we get
$\Rightarrow\text{I}=\frac{1}{2}\Big[\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^\pi_0-\frac{1}{2}\bigg\{\Big[\frac{1}{2}\text{e}^{2\text{x}}\cos\Big(\frac{\pi}{4}+{\text{x}}\Big)\Big]^{\pi}_0\\+\frac{1}{2}\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\bigg\}$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^\pi_0-\frac{1}{4}\Big[\text{e}^{2\text{x}}\cos\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^{\pi}_0-\frac{1}{4}\text{I}$
$\Rightarrow\frac{5}{4}\text{I}=\frac{1}{2}\Big[\text{e}^{2\pi}\sin\Big(\pi+\frac{\pi}{4}\Big)-\sin\Big(\frac{\pi}{4}\Big)\Big]\\-\frac{1}{4}\Big[\text{e}^{2\pi}\cos\Big(\pi+\frac{\pi}{4}\Big)-\cos\Big(\frac{\pi}{4}\Big)\Big]$
$\Rightarrow\frac{5}{4}\text{I}=\frac{1}{2}\Big[-\text{e}^{2\pi}\times\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\Big]-\frac{1}{4}\Big[-\text{e}^{2\pi}\times\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\Big]$
$\Rightarrow\frac{5}{4}\text{I}=-\frac{1}{2\sqrt{2}}\text{e}^{2\pi}-\frac{1}{2\sqrt{2}}+\frac{1}{4\sqrt{2}}\text{e}^{2\pi}+\frac{1}{4\sqrt{2}}$
$\Rightarrow\text{I}=-\frac{1}{5\sqrt{2}}\big(\text{e}^{2\pi}+1\big)$
View full question & answer→Question 1505 Marks
If f is an integrable function such that f(2a - x) = f(x), then prove that:
$\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=2\int\limits^{\text{a}}_0\text{f(x)}\text{dx}$
AnswerLet $\text{I}=\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
By Additive property
$\text{I}=\int\limits^{\text{a}}_0\text{f(x)}\text{dx}+\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
Consider the integral $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
Let $\text{x}=2\text{a}-\text{t},$ then $\text{dx}=-\text{dt}$
When $\text{x}=\text{a},\text{ t}=\text{a},\text{ x}=2\text{x},\text{t}=0$
Hence $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=-\int\limits^{0}_\text{a}\text{f(2a}-\text{t})\text{dt}$
$=\int\limits_{0}^\text{a}\text{f(2a}-\text{t})\text{dt}$
$=\int\limits_{0}^\text{a}\text{f(2a}-\text{x})\text{dx}$ (Changeing the varible)
Therefore,
$\text{I}=\int\limits_{0}^\text{a}\text{f(x})\text{dx}+\int\limits_{0}^\text{a}\text{f(2a}-\text{x})\text{dx}$
$=\int\limits_{0}^\text{a}\text{f(x})\text{dx}+\int\limits_{0}^\text{a}\text{f(x})\text{dx}$ $\Bigg[\text{Given}\int\limits_{0}^\text{a}\text{f(x})\text{dx}+\int\limits_{0}^\text{a}\text{f(2a}-\text{x})\text{dx}\Bigg]$
$=2\int\limits_{0}^\text{a}\text{f(x})\text{dx}$
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