Question 513 Marks
Evaluate the following integrals:
$\int\limits^{\text{a}}_0\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\text{a}}_0\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\text{ dx}\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\text{a}}_0\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{a}-\text{x}}+\sqrt{\text{x}}}\text{ dx}$
$\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$\text{I}=\int\limits^{\text{a}}_0\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\text{a}}_0\frac{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\text{ dx}$
$=\int\limits^{\text{a}}_0\text{dx}=\big[\text{x}\big]^{\text{a}}_0=\text{a}$
Hence, $\text{I}=\frac{\text{a}}{2}$
View full question & answer→Question 523 Marks
Evaluate the following integrals:
$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(2\sin|\text{x}|+\cos|\text{x}|\big)\text{dx}$
Answer$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(2\sin|\text{x}|+\cos|\text{x}|\big)\text{dx}$
$=\int^{0}_{-\frac{\pi}{4}}\big(-2\sin\text{x}+\cos\text{x}\big)\text{dx}+\int_{0}^{\frac{\pi}{2}}\big(2\sin\text{x}+\cos\text{x}\big)\text{dx}$
$=\big[2\cos\text{x}+\sin\text{x}\big]^0_{-\frac{\pi}{4}}+\big[-2\cos\text{x}+\sin\text{x}\big]_0^{\frac{\pi}{2}}$
$=2+0-0+1+0+1+2-0$
$=6$
View full question & answer→Question 533 Marks
Evaluate the following integrals:
$\int^\limits6_{-6}\big|\text{x}+2\big|\text{dx}$
Answer$\int^\limits6_{-6}\big|\text{x}+2\big|\text{dx}$
We know that,
$|\text{x}+2|=\begin{cases}-(\text{x}+2),&-6\leq\text{x}\leq-2\\\text{x}+2,&-2<\text{x}\leq6\end{cases}$
$\therefore\ \text{I}=\int^\limits6_{-6}\big|\text{x}+2\big|\text{dx}$
$\Rightarrow\text{I}=\int^\limits{-2}_{-6}\big(\text{x}+2\big)\text{dx}+\int^\limits6_{-2}\big(\text{x}+2\big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{-\text{x}^2}{2}-2\text{x}\Big]^{-2}_{-6}+\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^6_{-2}$
$\Rightarrow\text{I}=-2+4+18-12+18+12-2+4$
$\Rightarrow\text{I}=40$
View full question & answer→Question 543 Marks
Evaluate the following integrals:
$\int\limits^{{\pi}}_{-\frac{\pi}{2}}\sin^{-1}(\sin\text{x})\text{dx}$
Answer$\int\limits^{{\pi}}_{-\frac{\pi}{2}}\sin^{-1}\text{x}(\sin\text{x})\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^{-1}(\sin\text{x})\text{dx}\int\limits^{{\pi}}_{\frac{\pi}{2}}\sin^{-1}(\sin\text{x})\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{x dx}+\int\limits^{{\pi}}_{\frac{\pi}{2}}(\pi-\text{x})\text{dx}$ $\Big[\frac{\pi}{2}\leq\text{x}\leq\pi\Rightarrow-\pi\leq-\text{x}\leq-\frac{\pi}{2}\Rightarrow0\leq\pi-\text{x}\leq\frac{\pi}{2}\Big]$
$=\Big[\frac{\text{x}^2}{2}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}+\bigg[\frac{(\pi-\text{x})}{2\times(-1)}\bigg]^{\pi}_{\frac{\pi}{2}}$
$=\frac{1}{2}\Big(\frac{\pi^2}{4}-\frac{\pi^2}{4}\Big)-\frac{1}{2}\Big(0-\frac{\pi^2}{4}\Big)$
$=0+\frac{\pi^2}{8}$
$=\frac{\pi^2}{8}$
View full question & answer→Question 553 Marks
Evaluate the following integrals:
$\int^\limits4_1\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}4\text{x}+3,&\text{if }\ 1\leq\text{x}\leq2\\3\text{x}+5,&\text{if }\ 2\leq\text{x}\leq4\end{cases}$
AnswerWe have,
$\int^\limits4_1\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}4\text{x}+3,&\text{if }\ 1\leq\text{x}\leq2\\3\text{x}+5,&\text{if }\ 2\leq\text{x}\leq4\end{cases}$
$\text{I}=\int^\limits4_1\text{f(x)}\text{dx}$
$\Rightarrow\text{I}=\int^\limits2_1\text{f(x)}\text{dx}+\int^\limits4_2\text{f(x)}\text{dx}$ [Additive property]
$\Rightarrow\text{I}=\int^\limits2_1(4\text{x}+3)\text{dx}+\int^\limits4_2(3\text{x}+5)\text{dx}$
$\Rightarrow\text{I}=\Big[2\text{x}^2+3\text{x}\Big]^2_1+\Big[\frac{3\text{x}^2}{2}+5\text{x}\Big]^4_2$
$\Rightarrow\text{I}=8+6-2-3+24+20-6-10$
$\Rightarrow\text{I}=37$
View full question & answer→Question 563 Marks
Evaluate the following integrals:
$\int^\limits9_0\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}\sin\text{x},&0\leq\text{x}\leq\frac{\pi}{2}\\1,&\frac{\pi}{2}\leq\text{x}\leq3\\\text{e}^{\text{x}-3},&3\leq\text{x}\leq9\end{cases}$
AnswerWe have,
$\int^\limits9_0\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}\sin\text{x},&0\leq\text{x}\leq\frac{\pi}{2}\\1,&\frac{\pi}{2}\leq\text{x}\leq3\\\text{e}^{\text{x}-3},&3\leq\text{x}\leq9\end{cases}$
$\text{I}=\int^\limits9_0\text{f(x)}\text{dx}$
$\Rightarrow\text{I}=\int^\limits{\frac{\pi}{2}}_1\text{f(x)}\text{dx}+\int^\limits3_\frac{\pi}{2}\text{f(x)}\text{dx}+\int^\limits9_3\text{e}^{\text{x}-3}\text{ dx}$ [Additive property]
$\Rightarrow\text{I}=\int^\limits{\frac{\pi}{2}}_1\sin\text{x dx}+\int^\limits3_\frac{\pi}{2}\text{1 }\text{dx}+\int^\limits9_3\text{e}^{\text{x}-3}\text{ dx}$
$\Rightarrow\text{I}=\big[-\cos\text{x}\big]^{\frac{\pi}{2}}_0+\big[\text{x}\big]^3_\frac{\pi}{2}+\text{e}^6-\text{e}^0$
$\Rightarrow\text{I}=0+1+3-\frac{\pi}{2}+\text{e}^6-\text{e}^0$
$\Rightarrow\text{I}=3-\frac{\pi}{2}+\text{e}^6$
View full question & answer→Question 573 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{4}}_{\frac{-\pi}{4}}|\sin\text{x}|\text{dx}$
Answer$\int^\limits{\frac{\pi}{4}}_{\frac{-\pi}{4}}|\sin\text{x}|\text{dx}$
We know that,
$|\sin\text{x}|=\begin{cases}-\sin\text{x},&-\frac{\pi}{4}\leq\text{x}\leq0\\\sin\text{x},&0<\text{x}\leq\frac{\pi}{4}\end{cases}$
$\therefore\ \text{I}=\int^\limits{\frac{\pi}{4}}_{\frac{-\pi}{4}}|\sin\text{x}|\text{dx}$
$\Rightarrow\text{I}=\int^\limits0_{-\frac{\pi}{4}}-\sin\text{x dx}+\int\limits^{\frac{\pi}{4}}_0\sin\text{x dx}$
$\Rightarrow\text{I}=\big[\cos\text{x}\big]^0_{\frac{-\pi}{4}}-\big[\cos\text{x}\big]^{\frac{-\pi}{4}}_0$
$\Rightarrow\text{I}=1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+1$
$\Rightarrow\text{I}=2-\frac{2}{\sqrt{2}}$
$\Rightarrow\text{I}=2-\sqrt{2}$
View full question & answer→Question 583 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\text{e}^{\text{x}}(\sin\text{x}-\cos\text{x})\text{dx}$
Answer$\int\limits^{\frac{\pi}{2}}_0\text{e}^{\text{x}}(\sin\text{x}-\cos\text{x})\text{dx}$
$=-\int\limits^{\frac{\pi}{2}}_0\text{e}^{\text{x}}\Big[\cos\text{x}+(-\sin\text{x})\Big]\text{dx}$
$=-\big[\text{e}^{\text{x}}\cos\text{x}\big]^{\frac{\pi}{2}}_0$ $\Big\{\int\text{e}^{\text{x}}\big[\text{f(x)}+\text{f}'(\text{x})\big]\text{dx}=\text{e}^{\text{x}}\text{f(x)}+\text{C}\Big\}$
$=-\Big(\text{e}^{\frac{\pi}{2}}\cos\frac{\pi}{2}-\text{e}^0\cos0\Big)$
$=-\Big(\text{e}^{\frac{\pi}{2}}\times0-1\times1\Big)$
$=-(0-1)$
$=1$
View full question & answer→Question 593 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{2\pi}\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{2\pi}\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\text{dx}$ Then,
Integrating by parts,
$\text{I}=\Big[2\text{e}^{\text{x}}\sin\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0-\int_{0}^\limits{2\pi}2\text{e}^{\text{x}}\sin\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\text{dx}$
Integrating second term by parts,
$\text{I}=\Big[2\text{e}^{\text{x}}\sin\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0+\bigg\{\Big[4\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0\\+\int_{0}^\limits{2\pi}-4\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\text{ dx}\bigg\}$
$\Rightarrow\text{I}=\Big[2\text{e}^{\text{x}}\sin\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0+\Big[4\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0-4\text{I}$
$\Rightarrow5\text{I}=-2\text{e}^{2\pi}\frac{1}{\sqrt{2}}-2\frac{1}{\sqrt{2}}-4\text{e}^{2\pi}\frac{1}{\sqrt{2}}-4\frac{1}{\sqrt{2}}$
$\Rightarrow5\text{I}=-3\sqrt{2}\text{e}^{2\pi}-3\sqrt{2}$
$\Rightarrow\text{I}=-\frac{3\sqrt{2}}{5}\big(\text{e}^{2\pi}+1\big)$
View full question & answer→Question 603 Marks
Evaluate the following definite integrals:
$\int_{1}^\limits{\text{e}}\frac{\text{e}^{\text{x}}}{\text{x}}(1+\text{x}\log\text{x})\text{dx}$
AnswerLet $\text{I}=\int_{1}^\limits{\text{e}}\frac{\text{e}^{\text{x}}}{\text{x}}(1+\text{x}\log\text{x})\text{dx}$
$\text{I}=\int_{1}^\limits{\text{e}}\frac{\text{e}^{\text{x}}}{\text{x}}+\int_{1}^\limits{\text{e}}\text{e}^{\text{x}}\log\text{x}\text{ dx}$
$\text{I}=\big[\text{e}^{\text{x}}\log\text{x}\big]^{\text{e}}_1-\int_{1}^\limits{\text{e}}\text{e}^{\text{x}}\log\text{x}+\int_{1}^\limits{\text{e}}\text{e}^{\text{x}}\log\text{x}$
$\text{I}=\big[\text{e}^{\text{x}}\log\text{x}\big]^{\text{e}}_1$
$\text{I}=\big[\text{e}^{\text{x}}\log\text{e}-\text{e}^1\log1\big]$
$\text{I}=\big[\text{e}^{\text{e}}1-0\big]$
$\text{I}=\text{e}^{\text{e}}$
View full question & answer→Question 613 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\Big(\text{xe}^{2\text{x}}+\sin\frac{\pi\text{x}}{2}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\Big(\text{xe}^{2\text{x}}+\sin\frac{\pi\text{x}}{2}\Big)\text{dx}$ Then,
$\text{I}=\int_{0}^\limits{1}\text{xe}^{2\text{x}}\text{ dx}+\int_{0}^\limits{1}\sin\frac{\pi\text{x}}{2}\text{ dx}$
Integrating first term by parts,
$\text{I}=\Big[\text{x }\frac{\text{e}^{2\text{x}}}{2}\Big]^1_0-\int_{0}^\limits{1}1\frac{\text{e}^{2\text{x}}}{2}\text{ dx}+\bigg[-\frac{\cos\frac{\pi\text{x}}{2}}{\frac{\pi}{2}}\bigg]_0^1$
$\Rightarrow\text{I}=\Big[\text{x }\frac{\text{e}^{2\text{x}}}{2}\Big]^1_0-\Big[\frac{\text{e}^{2\text{x}}}{4}\Big]^1_0-\frac{2}{\pi}\Big[\cos\frac{\pi\text{x}}{2}\Big]^1_0$
$\Rightarrow\text{I}=\frac{\text{e}^{2}}{2}-\frac{\text{e}^{2}}{4}+\frac{1}{4}+\frac{2}{\pi}$
$\Rightarrow\text{I}=\frac{\text{e}^{2}}{4}+\frac{1}{4}+\frac{2}{\pi}$
View full question & answer→Question 623 Marks
Evaluate the following integrals:
$\int\limits^\text{b}_{\text{a}}\frac{\text{f(x)}}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\text{ dx}$
AnswerLet $\text{I}=\int\limits^\text{b}_{\text{a}}\frac{\text{f(x)}}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\text{ dx}\ ....(\text{i})$
$=\int\limits^\text{b}_{\text{a}}\frac{\text{f}(\text{a}+\text{b}-\text{x})}{\text{f}(\text{a}+\text{b}-\text{x})+\text{f}(\text{a}+\text{b}-\text{a}-\text{b}+\text{x})}\text{ dx}$
$=\int\limits^\text{b}_{\text{a}}\frac{\text{f}(\text{a}+\text{b}-\text{x})}{\text{f}(\text{a}+\text{b}-\text{x})+\text{f(x)}}\text{ dx}$
$\therefore\ \text{I}=\int\limits^\text{b}_{\text{a}}\frac{\text{f}(\text{a}+\text{b}-\text{x})}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^\text{b}_{\text{a}}\bigg[\frac{\text{f(x)}}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}+\frac{\text{f}(\text{a}+\text{b}-\text{x})}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\bigg]\text{dx}$
$=\int\limits^\text{b}_{\text{a}}\frac{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\text{ dx}$
$=\big[\text{x}\big]^{\text{b}}_\text{a}$
$=\text{b}-\text{a}$
Hence, $\text{I}=\frac{\text{b}-\text{a}}{2}$
View full question & answer→Question 633 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{4}\frac{1}{\sqrt{4\text{x}-\text{x}^2}}\text{ dx}$
AnswerWe have,
$\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{4\text{x}-\text{x}^2}}$
$=\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{4-4+4\text{x}-\text{x}^2}}$ [Add and subtract 4 in denominator]
$=\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{4-(\text{x}^2-4\text{x}+4)}}$
$=\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{(2)^2-(\text{x}-2)^2}}$
$=\Big[\sin^{-1}\Big(\frac{\text{x}-2}{2}\Big)\Big]^4_0$ $\bigg[\because\int\frac{\text{dx}}{\sqrt{\text{a}^2-\text{x}^2}}=\sin^{-1}\frac{\text{x}}{\text{a}}\bigg]$
$=\sin^{-1}(1)-\sin^{-1}(-1)$
$=\frac{\pi}{2}-\Big(-\frac{\pi}{2}\Big)$
$=\frac{2\pi}{2}=\pi$
View full question & answer→Question 643 Marks
Evaluate the following integrals:
$\int^\limits1_0\frac{1-\text{x}^2}{(1+\text{x}^2)^2}\text{ dx}$
AnswerLet $\text{I}=\int^\limits1_0\frac{1-\text{x}^2}{(1+\text{x}^2)^2}\text{ dx}$ Then,
$\text{I}=\int^\limits1_0\frac{\big(\frac{1}{\text{x}^2}-1\big)}{\big(\text{x}+\frac{1}{\text{x}}\big)^2}\text{ dx}$
Let $\text{x}+\frac{1}{\text{x}}=\text{t}$ Then, $1-\frac{1}{\text{x}^2}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=\infty$ and $\text{x}=1,\text{t}=2$
$\therefore\ \text{I}=\int^\limits2_\infty\frac{-\text{dt}}{\text{t}^2}$
$\Rightarrow\text{I}=\Big[\frac{1}{\text{t}}\Big]^2_\infty$
$\Rightarrow\text{I}=\frac{1}{2}-0$
$\Rightarrow\text{I}=\frac{1}{2}$
View full question & answer→Question 653 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\theta}{\sqrt{1+\cos\theta}}\text{ d}\theta$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\theta}{\sqrt{1+\cos\theta}}\text{ d}\theta$
Let $\cos\theta=\text{t}$ Then, $-\sin\theta\text{ d}\theta=\text{dt}$
When $\theta=0,\text{t}=1$ and $\theta=\frac{\pi}{2},\text{t}=0$
$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\theta}{\sqrt{1+\cos\theta}}\text{ d}\theta$
$=\int_{1}^\limits{0}\frac{-\text{dt}}{\sqrt{1+\text{t}}}$
$=\int_{0}^\limits{1}\frac{\text{dt}}{\sqrt{1+\text{t}}}$
$=2\big[\sqrt{1+\text{t}}\big]^1_0$
$=2\big(\sqrt{2}-1\big)$
View full question & answer→Question 663 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\text{x}(1-\text{x})^5\text{ dx}$
AnswerWe have,
$\int_{0}^\limits{1}\text{x}(1-\text{x})^5\text{ dx}$
Expanding (1 - x)5 by Binomial theorem,
$\therefore\ (1-\text{x})^5=1^5+{^5\text{C}_1}(-\text{x})+{^5\text{C}_2}(-\text{x})^2\\+{^5\text{C}_3}(-\text{x})^3+{^5\text{C}_4}(-\text{x})^4+{^5\text{C}_5}(-\text{x})^5$
$=1-5\text{x}+10\text{x}^2-10\text{x}^3+5\text{x}^4-\text{x}^5$
$=\int_{0}^\limits{1}\text{x}(1-5\text{x}+10\text{x}^2-10\text{x}^3+5\text{x}^4-\text{x}^5)\text{dx}$
$=\Big[\frac{\text{x}^2}{2}-\frac{5\text{x}^3}{3}+\frac{10\text{x}^4}{4}-\frac{10\text{x}^5}{5}+\frac{5\text{x}^6}{6}-\frac{\text{x}^7}{7}\Big]^1_0$
$=\frac{1}{2}-\frac{5}{3}+\frac{10}{4}-\frac{10}{5}+\frac{5}{6}-\frac{1}{7}$
$=\frac{1}{42}$
View full question & answer→Question 673 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}\text{ dx}$
AnswerLet $\text{e}^\text{x}=\text{t}$ Then, $\text{e}^\text{x}\text{ dx}=\text{dt}$
When $\text{e}^\text{x}=0,\text{t}=1$ and $\text{x}=1,\text{t}=\text{e}$
$\therefore\ \text{I}=\int_{0}^\limits{1}\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{1}^\limits{\text{e}}\frac{\text{dt}}{1+\text{t}^{2}}$
$\Rightarrow\text{I}=\big[\tan^{-1}\text{x}\big]^\text{e}_1$
$\Rightarrow\text{I}=\tan^{-1}\text{e}-\tan^{-1}1$
$\Rightarrow\text{I}=\tan^{-1}\text{e}-\frac{\pi}{4}$
View full question & answer→Question 683 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\text{xe}^{\text{x}^2}\text{ dx}$
AnswerLet $\text{x}^2=\text{t}$
Differentiating w.r.t. x, we get
$2\text{xdx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=1\Rightarrow\text{t}=1$
$\therefore\ \int_{0}^\limits{1}\text{xe}^{\text{x}^2}\text{ dx}=\int_{0}^\limits{1}\frac{\text{e}^{\text{t}}\text{ dt}}{2}$
$=\frac{1}{2}\int_{0}^\limits{1}\text{e}^{\text{t}}\text{ dt}$
$=\frac{1}{2}\big[\text{e}^{\text{t}}\big]^1_0$
$=\frac{1}{2}\big[\text{e}^1-\text{e}_0\big]$ $\big[\because\text{e}^0=1\big]$
$=\frac{1}{2}\big(\text{e}-1\big)$
View full question & answer→Question 693 Marks
Evaluate the following definite integrals:
$\int_{\frac{\pi}{4}}^\limits{\frac{\pi}{2}}\cot\text{x}\text{ dx}$
AnswerWe have,
$\int_{\frac{\pi}{4}}^\limits{\frac{\pi}{2}}\cot\text{x}\text{ dx}$
We know that $\int\cot\text{x dx}=\log(\sin\text{x})$
Now, $\int_{\frac{\pi}{4}}^\limits{\frac{\pi}{2}}\cot\text{x}\text{ dx}$
$=\big[\log(\sin\text{x})\big]^{\frac{\pi}{2}}_\frac{\pi}{4}$
$=\Big[\log\Big(\sin\frac{\pi}{2}\Big)-\log\Big(\sin\frac{\pi}{4}\Big)\Big]$
$=\Big[\log1-\log\frac{1}{\sqrt{2}}\Big]$
$=\big[0-\log\text{2}\big]$
$=\log\sqrt{2}$ $[\because\log\text{a}^{\text{n}}-\text{n}\log\text{a}\big]$
$=\frac{1}{2}\log2$
View full question & answer→Question 703 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}$
AnswerWe have,
$\text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}$
Let $\text{f}(\text{x})=\text{x}\cos^2\text{x}$
$\Rightarrow \text{f}(-\text{x})= (-\text{x})\cos^2(-\text{x})$
$= - \text{x}\cos^2\text{x}$
$\therefore \text{f}(-\text{x})=-\text{f}(\text{x})$
i.e., f(x) is odd function.
We know that $\int\limits^\text{a}_{-\text{a}}\text{f}(\text{x})\text{dx} = 0, $ if f(x) is odd function.
$\therefore\ \text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}=0$
View full question & answer→Question 713 Marks
Evaluate the following integrals:
$\int_{2}^\limits{4}\frac{\text{x}}{\text{x}^2+1}\text{ dx}$
AnswerLet $\text{x}^2=\text{t}$ Then, $2\text{x dx}=\text{dt}$
When $\text{x}=2,\text{ t}=4$ and $\text{x}=4,\text{ t}=16$
$\therefore\ \text{I}=\int_{2}^\limits{4}\frac{\text{x}}{\text{x}^2+1}\text{ dx}$
$\Rightarrow\text{I}=\int_{4}^\limits{16}\frac{1}{2}\frac{\text{dt}}{\text{t}+1}$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\log(\text{t}+1)\Big]^{16}_4$
$\Rightarrow\text{I}=\frac{1}{2}\log17-\frac{1}{2}\log5$
$\Rightarrow\text{I}=\frac{1}{2}\log\frac{17}{5}$
View full question & answer→Question 723 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\log\tan\text{x dx}$
AnswerLet, $\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\tan\text{x dx}\ ....(\text{i})$ $=\int\limits^{\frac{\pi}{2}}_0\log\tan\Big(\frac{\pi}{2}-\text{x}\Big)\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$ $=\int\limits^{\frac{\pi}{2}}_0\log\cot\text{x dx}\ ...(\text{ii})$ Adding (i) and (ii) we get $2\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\tan\text{x dx}+\int\limits^{\frac{\pi}{2}}_0\log\cot\text{x dx}$ $=\int\limits^{\frac{\pi}{2}}_0\log(\tan\text{x}\cdot\cot\text{x})\text{ dx}$ $=\int\limits^{\frac{\pi}{2}}_0\log1\text{ dx}=0$ Hence, $\text{I}=0$
View full question & answer→Question 733 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\pi}\frac{1}{1+\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\pi}\frac{1}{1+\sin\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\pi}\frac{1-\sin\text{x}}{(1+\sin\text{x})(1-\sin\text{x})}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\pi}\frac{1-\sin\text{x}}{1+\sin^2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\pi}\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{ dx}$ $\big[\because\sin^2\text{x}+\cos^2\text{x}=1\big]$
$\Rightarrow\text{I}=\int_{0}^\limits{\pi}\sec^2\text{x}-\sec\text{x}\tan\text{x}\text{ dx}$
$\Rightarrow\text{I}=\big[\tan\text{x}-\sec\text{x}\big]^{\pi}_0$
$\Rightarrow\text{I}=(\tan\pi-\sec\pi)-(\tan0-\sec0)$
$\Rightarrow\text{I}=0+1-(0-1)$
$\Rightarrow\text{I}=1+1$
$\Rightarrow\text{I}=2$
View full question & answer→Question 743 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos2\text{x}\text{ dx}$ Then,
Integrating by parts
$\text{I}=\Big[\text{x}^2\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\int_{0}^\limits{\frac{\pi}{2}}2\text{x}\frac{\sin2\text{x}}{2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\text{x}^2\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\Big[-\text{x}\frac{\cos2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0+\int_{0}^\limits{\frac{\pi}{2}}-1\frac{\cos2\text{x}}{2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\text{x}^2\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\Big[-\text{x}\frac{\cos2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\Big[\frac{\sin2\text{x}}{4}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0-\frac{\pi}{4}-0$
$\Rightarrow\text{I}=-\frac{\pi}{4}$
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