If I_{1}=int_{0}^{pi / 2} x sin x , dx and I_ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We are given $I_{1} = \int_{0}^{\pi / 2} x \sin x \, dx$ and $I_{2} = \int_{0}^{\pi / 2} x \cos x \, dx$. Using integration by parts,$\int u \, dv

Vedclass · Accepted Answer

$I_{1}+I_{2}=\frac{\pi}{2}$

Vedclass · Answer

(D) We are given $I_{1} = \int_{0}^{\pi / 2} x \sin x \, dx$ and $I_{2} = \int_{0}^{\pi / 2} x \cos x \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$. For $I_{1}$,let $u = x$ and $dv = \sin x \, dx$. Then $du = dx$ and $v = -\cos x$. $I_{1} = [x(-\cos x)]_{0}^{\pi / 2} - \int_{0}^{\pi / 2} (-\cos x) \, dx = [0 - 0] + [\sin x]_{0}^{\pi / 2} = 1 - 0 = 1$. For $I_{2}$,let $u = x$ and $dv = \cos x \, dx$. Then $du = dx$ and $v = \sin x$. $I_{2} = [x \sin x]_{0}^{\pi / 2} - \int_{0}^{\pi / 2} \sin x \, dx = [\frac{\pi}{2} \sin(\frac{\pi}{2}) - 0] - [-\cos x]_{0}^{\pi / 2} = \frac{\pi}{2} - [-(0 - 1)] = \frac{\pi}{2} - 1$. Adding the two results,$I_{1} + I_{2} = 1 + (\frac{\pi}{2} - 1) = \frac{\pi}{2}$. Thus,the correct option is $D$.

If $I_{1}=\int_{0}^{\pi / 2} x \sin x \, dx$ and $I_{2}=\int_{0}^{\pi / 2} x \cos x \, dx$,then which one of the following is true?

Similar Questions

Evaluate the definite integral $\int_{0}^{9} [\sqrt{x} + 2] \, dx$,where $[\cdot]$ denotes the Greatest Integer Function $(G.I.F.)$.

$\int_0^1 {{\cos }^{ - 1}}x\,dx = $

The value of the definite integral $\int_{1}^{\infty} (e^{x+1} + e^{3-x})^{-1} \, dx$ is

Let $I_{1}=\int_{0}^{n}[x] d x$ and $I_{2}=\int_{0}^{n}\{x\} d x,$ where $[x]$ and $\{x\}$ are the integral and fractional parts of $x$ respectively, and $n \in N-\{1\}.$ Then, $I_{1} / I_{2}$ is equal to

$\int_0^4 |2x - 5| \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module