Question 513 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 523 Marks
Evaluate the following integrals:
$\int^\limits{2}_{-2}|\text{x}+1|\text{dx}$
Answer$\int^\limits{2}_{-2}|\text{x}+1|\text{dx}=\int^\limits{-1}_{-2}-(\text{x}+1)\text{dx}+\int^\limits{2}_{-1}(\text{x}+1)\text{dx}$
$=-\Big[\frac{\text{x}^2}{2}+\text{x}\Big]^{-1}_{-2}+\Big[\frac{\text{x}^2}{2}+\text{x}\Big]^2_{-1}$
$=-\bigg[\Big(\frac{1}{2}-1\Big)-\Big(\frac{4}{2}-2\Big)\bigg]+\bigg[\Big(\frac{4}{2}-2\Big)-\Big(\frac{1}{2}-1\Big)\bigg]$
$=-\bigg[\Big(-\frac{1}{2}\Big)-0\bigg]+\Big[4+\frac{1}{2}\Big]$
$=\frac{1}{2}+4\frac{1}{2}$
$=5$
View full question & answer→Question 533 Marks
Evaluate the following integrals:
$\int^\limits\frac{\pi}{4}_{0}\sin^32\text{t}\cos2\text{t}\text{ dt}$
AnswerLet $\text{I}\int^\limits\frac{\pi}{4}_{0}\sin^32\text{t}\cos2\text{t}\text{ dt}$ Then,Let $\sin2\text{t}=\text{u}$ Then, $2\cos2\text{t dt} =\text{du}$
When $\text{t}=0,\text{u}=0$ and $\text{t}=\frac{\pi}{4},\text{u}=1$
$\therefore\ \text{I}=\frac{1}{2}\int^\limits{1}_{0}\text{u}^3\text{ du}$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\frac{\text{u}^4}{4}\Big]^1_0$
$\Rightarrow\text{I}=\frac{1}{2}\Big(\frac{1}{4}-0\Big)$
$\Rightarrow\text{I}=\frac{1}{8}$
View full question & answer→Question 543 Marks
Evaluate the following integrals:
$\int^\limits3_{-3}|\text{x}+1|\text{dx}$
Answer$\text{I}=\int^\limits3_{-3}|\text{x}+1|\text{dx}$
We know that,
$|\text{x}+1|=\begin{cases}-(\text{x}+1),&-3\leq\text{x}\leq-1\\\text{x}+1,&-1<\text{x}\leq3\end{cases}$
$\therefore\ \text{I}=\int^\limits{-1}_{-3}-(\text{x}+1)\text{dx}+\int^\limits{3}_{-1}\big[\text{x}+1\big]\text{dx}$
$\Rightarrow\text{I}=\bigg[-\frac{(\text{x}+1)^2}{2}\bigg]+\bigg[\frac{(\text{x}+1)^2}{2}\bigg]^3_{-1}$
$\Rightarrow\text{I}=0+2+8-0$
$\Rightarrow\text{I}=10$
View full question & answer→Question 553 Marks
If $[\cdot]$ and $\{\cdot\}$ denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
$\int\limits^{\frac{\pi}{4}}_0\sin\{\text{x}\}\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{\frac{\pi}{4}}_0\sin\{\text{x}\}\text{dx}$
We know that,
$\{\text{x}\}=\text{x},\text{ when }0<\text{x}<\frac{\pi}{4}$ $\big(\text{As }\pi=3.14\Rightarrow\frac{\pi}{4}=0.784<1\big)$
$\therefore\ \text{I}=\int\limits^{\frac{\pi}{4}}_0\sin\{\text{x}\}\text{dx}$
$=\big[-\cos\text{x}\big]^{\frac{\pi}{4}}_{0}$
$=-\Big(\cos\frac{\pi}{4}-\cos\frac{\pi}{4}\Big)$
$=\cos0-\cos\frac{\pi}{4}$
$=1-\frac{1}{\sqrt{2}}$
$=\frac{\sqrt{2}-1}{\sqrt{2}}$
View full question & answer→Question 563 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
Answer$\int_{0}^\limits{1}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
$=\int_{0}^\limits{1}2\tan^{-1}\text{x dx}$
$=2\Big[\text{x}\tan^{-1}\text{x}\Big]^1_0-2\int_{0}^\limits{1}\frac{\text{x}}{1+\text{x}^2}\text{ dx}$
$=2\Big[\text{x}\tan^{-1}\text{x}\Big]^1_0-\Big[\log\big(1+\log\text{x}^2\big)\Big]$
$=2\frac{\pi}{4}-0-\log2+0$
$=\frac{\pi}{2}-\log2$
View full question & answer→Question 573 Marks
Evaluate the following integrals:
$\int^\limits{9}_4\frac{\sqrt{\text{x}}}{\big(30-\text{x}^{\frac{3}{2}}\big)^2}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{9}_4\frac{\sqrt{\text{x}}}{\big(30-\text{x}^{\frac{3}{2}}\big)^2}\text{ dx}$ Then,
Let $\Big(30-\text{x}^{\frac{3}{2}}\Big)=\text{t}$ Then, $-\frac{3}{2}\sqrt{\text{x}}\text{ dx}=\text{dt}$
When $\text{x}=4,\text{t}=22$ and $\text{x}=9,\text{t}=3$
$\therefore\ \text{I}=\int^\limits{3}_{22}-\frac{2}{3}\frac{1}{\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=\frac{2}{3}\Big[\frac{1}{\text{t}}\Big]^3_{22}$
$\Rightarrow\text{I}=\frac{2}{3}\Big(\frac{1}{3}-\frac{1}{22}\Big)$
$\Rightarrow\text{I}=\frac{19}{99}$
View full question & answer→Question 583 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 593 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 603 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 613 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{2}\frac{1}{\sqrt{3+2\text{x}-\text{x}^2}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{2}\frac{1}{\sqrt{3+2\text{x}-\text{x}^2}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{2}\frac{1}{\sqrt{-\text{x}^2+2\text{x}-1+1+3}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{2}\frac{1}{\sqrt{-(\text{x}-1)^2+4}}\text{ dx}$
$\Rightarrow\text{I}=\Big[\sin^{-1}\frac{(\text{x}-1)}{2}\Big]^{2}_0$
$\Rightarrow\text{I}=\sin^{-1}\frac{1}{2}-\sin^{-1}\Big(-\frac{1}{2}\Big)$
$\Rightarrow\text{I}=2\sin^{-1}\frac{1}{2}$
$\Rightarrow\text{I}=2\times\frac{\pi}{6}$
$\Rightarrow\text{I}=\frac{\pi}{3}$
View full question & answer→Question 623 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{2\text{x}}{1+\text{x}^4}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{2\text{x}}{1+\text{x}^4}\text{ dx}$
Let $\text{x}^2=\text{t}$ Then, $2\text{xdx}=\text{dt}$
When $\text{x}=10,\text{t}=0$ and $\text{x}=1,\text{t}=1$
$\therefore\ \text{I}=\int_{0}^\limits{1}\frac{2\text{x}}{1+\text{x}^4}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{1}\frac{1}{1+\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=\big[\tan^{-1}\text{t}\big]^1_0$
$\Rightarrow\text{I}=\tan^{-1}1-\tan^{-1}0$
$\Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 633 Marks
Evaluate the following integrals:
$\int^\limits{2}_{1}|\text{x}-3|\text{dx}$
Answer$\int^\limits{2}_{1}|\text{x}-3|\text{dx}$
We know that,
$|\text{x}+1|=\begin{cases}-(\text{x}+1),&1\leq\text{x}\leq3\$\text{x}+1),&\text{x}>3\end{cases}$
$\therefore\ \text{I}=\int^\limits{2}_{1}|\text{x}-3|\text{dx}$
$\Rightarrow\text{I}=\int^\limits{2}_{1}-(\text{x}-3)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{-\text{x}^2}{2}-3\text{x}\Big]^2_1$
$\Rightarrow\text{I}=-2-6+\frac{1}{2}+3$
$\Rightarrow\text{I}=\frac{3}{2}$
View full question & answer→Question 643 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
$\text{I}=\Big[\text{x}^2\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\int_{0}^\limits{\frac{\pi}{2}}2\text{x}\frac{\sin2\text{x}}{2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\text{x}^2\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\Big[-\text{x}\frac{\cos2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0+\int_{0}^\limits{\frac{\pi}{2}}-1\frac{\cos2\text{x}}{2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\text{x}^2\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\Big[-\text{x}\frac{\cos2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\Big[\frac{\sin2\text{x}}{4}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0-\frac{\pi}{4}-0$
$\Rightarrow\text{I}=-\frac{\pi}{4}$
View full question & answer→Question 653 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 663 Marks
Evaluate the following definite integrals:
$\int_{-1}^\limits{1}\frac{1}{\text{x}^2+2\text{x}+5}\text{ dx}$
AnswerLet $\text{I}=\int_{-1}^\limits{1}\frac{1}{\text{x}^2+2\text{x}+5}\text{ dx}$ Then,
$\text{I}=\int_{-1}^\limits{1}\frac{1}{\big(\text{x}^2+2\text{x}+1\big)+4}\text{ dx}$
$\Rightarrow\text{I}=\int_{-1}^\limits{1}\frac{1}{(\text{x}+1)^2+2^2}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\tan^{-1}\frac{(\text{x}+1)}{2}\Big]^1_{-1}$
$\Rightarrow\text{I}=\frac{1}{2}\Big(\frac{\pi}{4}\Big)$
$\Rightarrow\text{I}=\frac{\pi}{8}$
View full question & answer→Question 673 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 683 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 693 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 703 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 713 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 723 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 733 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 743 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 $\text{I}=\int_{1}^\limits{\text{e}}\frac{\text{e}^{\text{x}}}{\text{x}}(1+\text{x}\log\text{x})\text{dx}$
$\text{I}=\int_{1}^\limits{\text{e}}\frac{\text{e}^{\text{x}}}{\text{x}}+\int_{1}^\limits{\text{e}}\text{e}^{\text{x}}\log\text{x}\text{ dx}$
$\text{I}=\big[\text{e}^{\text{x}}\log\text{x}\big]^{\text{e}}_1-\int_{1}^\limits{\text{e}}\text{e}^{\text{x}}\log\text{x}+\int_{1}^\limits{\text{e}}\text{e}^{\text{x}}\log\text{x}$
$\text{I}=\big[\text{e}^{\text{x}}\log\text{x}\big]^{\text{e}}_1$
$\text{I}=\big[\text{e}^{\text{x}}\log\text{e}-\text{e}^1\log1\big]$
$\text{I}=\big[\text{e}^{\text{e}}1-0\big]$
$\text{I}=\text{e}^{\text{e}}$
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