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Maths Solve the Following Question.(5 Marks)

Definite Integration · Maharashtra Board STD 12 Science (MSBSHSE - English Medium). 13 questions.

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Solve the Following Question.(5 Marks)

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13 questions · self-marked practice — reveal the answer and mark yourself.

Question 35 Marks
Evaluate : $\int_{-\pi}^\pi \frac{x(1+\sin x)}{1+\cos ^2 x} \cdot d x$
Answer
Let $\begin{aligned} \mathrm{I} & =\int_{-\pi}^\pi \frac{x(1+\sin x)}{1+\cos ^2 x} \cdot d x \\ & =\left[\left(\int_{-\pi}^\pi \frac{x}{1+\cos ^2 x} \cdot d x\right)+\left(\int_{-\pi}^\pi \frac{x \cdot \sin x}{1+\cos ^2 x} \cdot d x\right)\right]\end{aligned}$
The function $\frac{x}{1+\cos ^2 x}$ is odd function and the function $\frac{x \cdot \sin x}{1+\cos ^2 x}$ is even function.
$
\begin{aligned}
& \int_{-a}^a f(x) d x=2 \cdot \int_0^a f(x) d x \text {, if } f(x) \text { even function } \\
& =0 \quad, \text { if } f(x) \text { is odd function } \\
& \therefore \mathrm{I}=0+2 \cdot \int_0^\pi \frac{x \cdot \sin x}{1+\cos ^2 x} \cdot d x \\
& \therefore \mathrm{I}=2 \cdot \int_0^\pi \frac{x \cdot \sin x}{1+\cos ^2 x} \cdot d x \\
& =2 \cdot \int_0^\pi \frac{(\pi-x) \cdot \sin (\pi-x)}{1+[\cos (\pi-x)]^2} \cdot d x \\
& =2 \cdot \int_0^\pi \frac{(\pi-x) \cdot \sin x}{1+(-\cos x)^2} \cdot d x \\
& =2 \pi \cdot \int_0^\pi \frac{\pi \cdot \sin x-x \cdot \sin x}{1+\cos ^2 x} \cdot d x \\
& =2 \pi \cdot \int_0^\pi \frac{\sin x}{1+\cos ^2 x}-2 \cdot \int_0^\pi \frac{x \cdot \sin x}{1+\cos ^2 x} \cdot d x \\
& \mathrm{I}=2 \pi \cdot \int_0^\pi \frac{\sin x}{1+\cos ^2 x}-\mathrm{I} \quad \ldots \text { by eq.(i) } \\
& \mathrm{I}+\mathrm{I}=2 \pi \cdot \int_0^\pi \frac{\sin x}{1+\cos ^2 x} \\
& \text { put } \cos x=t \quad \therefore \quad-\sin x \cdot d x=+d t \\
\end{aligned}
$
As varies from 0 to $\pi, t$ varies from 1 to -1
$
\begin{aligned}
& 2 \mathrm{I}=2 \pi \cdot \int_{-1}^1 \frac{-1}{1+t^2} \cdot d t \\
& \mathrm{I}=\pi \cdot 2 \int_0^1 \frac{1}{1+t^2} \cdot d t \quad \text { (where } \frac{1}{1+t^2} \text { is even function.) }
\end{aligned}
$

$\begin{aligned} & \mathrm{I}=2 \pi \cdot\left[\tan ^{-1} t\right]_0^1 . \\ & =2 \pi\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right] \\ & =2 \pi\left(\frac{\pi}{4}-0\right)=\frac{\pi^2}{2} \\ & \therefore \quad \int_{-\pi}^\pi \frac{x(1+\sin x)}{1+\cos ^2 x} \cdot d x=\frac{\pi^2}{2} \\ & \end{aligned}$

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Question 55 Marks
Evaluate : $\int_0^2 \frac{2^x}{2^x\left(1+4^x\right)} \cdot d x$
Answer
Let $I=\int_0^2 \frac{2^x}{2^x\left(1+4^x\right)} \cdot d x$
put $2^x=t \quad \therefore \quad 2^x \cdot \log 2 \cdot d x=1 \cdot d t$
As $x$ varies from 0 to $2, t$ varies from 1 to 4
$
\begin{aligned}
& =\int_1^4 \frac{\frac{1}{\log 2}}{t\left(1+t^2\right)} \cdot d t \\
& =\frac{1}{\log 2} \cdot \int_1^4 \frac{1}{t\left(1+t^2\right)} \cdot d t \\
& =\frac{1}{\log 2} \cdot \int_1^4 \frac{1+t^2-t^2}{t\left(1+t^2\right)} \cdot d t
\end{aligned}
$
may be solved by method of partial fraction
$
\begin{aligned}
& =\frac{1}{\log 2} \cdot \int_1^4\left[\frac{1+t^2}{t\left(1+t^2\right)}-\frac{t^2}{t\left(1+t^2\right)}\right] \cdot d t \\
& =\frac{1}{\log 2} \cdot \int_1^4\left[\frac{1}{t}-\frac{t}{1+t^2}\right] \cdot d t \\
& =\frac{1}{\log 2} \cdot\left[\int_1^4 \frac{1}{t} \cdot d t-\frac{1}{2} \int_1^4 \frac{2 t}{1+t^2} \cdot d t\right]
\end{aligned}
$
$\begin{aligned} & =\frac{1}{\log 2} \cdot\left[\log (t)-\frac{1}{2} \log \left(1+t^2\right)\right]_1^4 \\ & =\frac{1}{\log 2} \cdot\left[\left(\log 4-\frac{1}{2} \log 17\right)-\right. \\ & =\frac{1}{\log 2} \cdot\left[\log 4-\frac{1}{2} \log 17+\frac{1}{2} \log 2\right] \\ & \because \log 1=0 \\ & =\frac{1}{\log 2} \cdot\left[\log \frac{4 \sqrt{2}}{\sqrt{17}}\right] \\ & \therefore \int_0^2 \frac{2^x}{2^x\left(1+4^x\right)} \cdot d x=\frac{1}{(\log 2)} \cdot\left[\log \frac{4 \sqrt{2}}{\sqrt{17}}\right] \\ & =\log _2\left(\frac{4 \sqrt{2}}{\sqrt{17}}\right)\end{aligned}$
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Question 65 Marks
Evaluate : $\int_0^{\pi / 2} \sin x \cdot d x$
Answer
$
\begin{aligned}
& \int_0^{\pi / 2} \sin x \cdot d x=\int_0^{\pi / 2} f(x) d x \\
& f(x)=\sin x \quad a=0 ; b=\frac{\pi}{2} \\
& \Rightarrow \quad f(a+r h)=\sin (a+r h) \\
& =\sin (0+r h) \\
& =\sin r h \\
& h=\frac{b-a}{n}=\frac{\frac{\pi}{2}-0}{n} \\
& \therefore \quad n h=\frac{\pi}{2} \\
&
\end{aligned}
$
and
We know $\int_a^b f(x) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot[f(a+r h)]$
$
\begin{aligned}
\therefore \quad \int_0^{\pi / 2} \sin x \cdot d x & =\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot \sin r h \\
& =\lim _{n \rightarrow \infty} h \cdot \sum_{r=1}^n \sin r h \\
& =\lim _{n \rightarrow \infty} h \cdot[\sin h+\sin 2 h+\sin 3 h+\ldots+\sin n h]
\end{aligned}
$
Consider,
$
\begin{aligned}
& \sum_{r=1}^n \sin r h=\sin h+\sin 2 h+\sin 3 h+\ldots+\sin n h \\
& =2 \sin \frac{h}{2} \cdot \sin h+2 \sin \frac{h}{2} \cdot \sin 2 h+2 \sin \frac{h}{2} \cdot \sin 3 h+\ldots+2 \sin \frac{h}{2} \cdot \sin n h \\
& \because \quad 2 \sin \mathrm{A} \cdot \sin \mathrm{B}=\cos (\mathrm{A}-\mathrm{B})-\cos (\mathrm{A}+\mathrm{B}) \\
& 2 \sin \frac{h}{2} \cdot \sum_{r=1}^n \sin r h=\left[\left(\cos \frac{h}{2}-\cos \frac{3 h}{2}\right)+\left(\cos \frac{3 h}{2}-\cos \frac{5 h}{2}\right)+\left(\cos \frac{5 h}{2}-\cos \frac{7 h}{2}\right)+\ldots\right. \\
& +\ldots+\left(\cos \left(\frac{2 n-1}{2}\right) h-\left(\cos \left(\frac{2 n+1}{2}\right) h\right]\right. \\
& =\left[\cos \frac{h}{2}-\cos \left(\frac{2 n+1}{2}\right) h\right] \\
& =\left[\cos \frac{h}{2}-\cos \left(\frac{2 n h}{2}+\frac{h}{2}\right)\right] \\
& =\left[\cos \frac{h}{2}-\cos \left(\frac{\pi}{2}+\frac{h}{2}\right)\right] \\
& \because \quad n h=\frac{\pi}{2} \\
& =\left(\cos \frac{h}{2}+\sin \frac{h}{2}\right) \\
&
\end{aligned}
$

$\begin{aligned} & \therefore \quad \sum_{r=1}^n \sin r h=\frac{\cos \frac{h}{2}+\sin \frac{h}{2}}{2 \sin \frac{h}{2}} \\ & \text { Now from I, } \\ & \int_0^{\pi / 2} \sin x \cdot d x=\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot \sin r h \\ & =\lim _{n \rightarrow \infty} h \cdot\left[\frac{\cos \frac{h}{2}+\sin \frac{h}{2}}{2 \sin \frac{h}{2}}\right] \\ & \because \quad n h=\frac{\pi}{4} \text { as } n \rightarrow \infty \Rightarrow h \rightarrow 0\left(\frac{1}{n} \rightarrow 0\right) \\ & =\lim _{\substack{h \rightarrow \infty \\ h \rightarrow 0}}\left[\frac{\cos \frac{h}{2}+\sin \frac{h}{2}}{\frac{2 \cdot \sin \frac{h}{2}}{h}}\right] \\ & =\frac{\cos 0+\sin 0}{\left(\frac{1}{2}\right)} \\ & =\frac{1+0}{2 \cdot \frac{1}{2}}=1 \\ & \therefore \quad \int_0^{\pi / 2} \sin x \cdot d x=1 \\ & \end{aligned}$

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Question 75 Marks
Evaluate : $\int_0^4\left(x-x^2\right) \cdot d x$
Answer
$
\begin{aligned}
& \int_0^4\left(x-x^2\right) \cdot d x=\int_a^b f(x) d x \\
& f(x)=x-x^2 \quad a=0 ; b=4 \\
& \Rightarrow \quad f(a+r h)=f(0+r h) \quad \text { and } \\
& =f(r h) \\
& =(r h)-(r h)^2 \\
& h=\frac{b-a}{n} \\
& =r h-r^2 h^2 \\
& h=\frac{4-0}{n} \\
& \therefore \quad n h=4 \\
&
\end{aligned}
$
We know $\int_a^b f(x) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot[f(a+r h)]$
$
\begin{aligned}
\therefore \quad \int_0^4\left(x-x^2\right) \cdot d x & =\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot\left(r h-r^2 h^2\right) \\
& =\lim _{n \rightarrow \infty} \sum_{r=1}^n\left(r h^2-r^2 h^3\right) \\
& =\lim _{n \rightarrow \infty}\left(h^2 \cdot \sum_{r=1}^n r-h^3 \cdot \sum_{r=1}^n r^2\right) \\
& =\lim _{n \rightarrow \infty}\left[h^2\left(\frac{n(n+1)}{2}\right)-h^3\left(\frac{n(n+1)(2 n+1)}{6}\right)\right] \\
& =\lim _{n \rightarrow \infty}\left[\frac{h^2 \cdot n \cdot n\left(1+\frac{1}{n}\right)}{2}-\frac{h^3 \cdot n \cdot n\left(1+\frac{1}{n}\right) n\left(2+\frac{1}{n}\right)}{6}\right] \\
& =\lim _{n \rightarrow \infty}\left[\frac{(n h)^2\left(1+\frac{1}{n}\right)}{2}-\frac{(n h)^3\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}\right] \\
& =\lim _{n \rightarrow \infty}\left[\frac{(4)^2\left(1+\frac{1}{n}\right)}{2}-\frac{(4)^3\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}\right] \\
& =\frac{(4)^2 \cdot(1+0)}{2}-\frac{(4)^3(1+0)(2+0)}{6} \\
& =-\frac{(64)(2)}{6}
\end{aligned}
$
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Solve the Following Question.(5 Marks) - Maths STD 12 Science Questions - Vidyadip