The value of int_{ - 1}^1 {frac{{sin x - {x^2 | vedclass

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

Question

(C) Let $I = \int_{ - 1}^1 {\frac{{\sin x - {x^2}}}{{3 - |x|}}} \,dx$. We can split the integral as: $I = \int_{ - 1}^1 {\frac{{\sin x}}{{3 - |x|}}} \

Vedclass · Accepted Answer

$2\int_0^1 {\frac{{ - {x^2}}}{{3 - |x|}}} \,dx$

Vedclass · Answer

(C) Let $I = \int_{ - 1}^1 {\frac{{\sin x - {x^2}}}{{3 - |x|}}} \,dx$. We can split the integral as: $I = \int_{ - 1}^1 {\frac{{\sin x}}{{3 - |x|}}} \,dx - \int_{ - 1}^1 {\frac{{{x^2}}}{{3 - |x|}}} \,dx$. Let $f(x) = \frac{{\sin x}}{{3 - |x|}}$ and $g(x) = \frac{{{x^2}}}{{3 - |x|}}$. For $f(x)$,$f(-x) = \frac{{\sin(-x)}}{{3 - |-x|}} = \frac{{-\sin x}}{{3 - |x|}} = -f(x)$,so $f(x)$ is an odd function. Thus,$\int_{ - 1}^1 {\frac{{\sin x}}{{3 - |x|}}} \,dx = 0$. For $g(x)$,$g(-x) = \frac{{(-x)^2}}{{3 - |-x|}} = \frac{{{x^2}}}{{3 - |x|}} = g(x)$,so $g(x)$ is an even function. Thus,$\int_{ - 1}^1 {\frac{{{x^2}}}{{3 - |x|}}} \,dx = 2\int_0^1 {\frac{{{x^2}}}{{3 - |x|}}} \,dx$. Substituting these back into $I$: $I = 0 - 2\int_0^1 {\frac{{{x^2}}}{{3 - |x|}}} \,dx = 2\int_0^1 {\frac{{ - {x^2}}}{{3 - |x|}}} \,dx$.

The value of $\int_{ - 1}^1 {\frac{{\sin x - {x^2}}}{{3 - |x|}}\,dx} $ is

Similar Questions

Let $f: R \rightarrow R$ be a continuous function such that $f(x)+f(x+1)=2$ for all $x \in R$. If $I_{1}=\int_{0}^{8} f(x) d x$ and $I_{2}=\int_{-1}^{3} f(x) d x$,then the value of $I_{1}+2 I_{2}$ is equal to:

$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt[3]{\sec x}}{\sqrt[3]{\sec x}+\sqrt[3]{\operatorname{cosec} x}} d x=$

The value of $\int_0^{n\pi + v} {|\sin x|\,dx} $ is

If $S_n = \int_0^{\frac{\pi}{2}} \frac{\sin((2n-1)x)}{\sin x} dx$ and $n$ is an integer,then $S_{n+1} - S_n =$

$\int_0^\pi \frac{x \tan x}{\sec x \cdot \operatorname{cosec} x} d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module