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

If $\int \limits_0^\pi \frac{5^{\cos x}(1+\cos x \cos 3x+\cos^2 x+\cos^3 x \cos 3x) dx}{1+5^{\cos x}} = \frac{k \pi}{16}$,then $k$ is equal to $...........$.

If $b = \int_{0}^{1} \frac{e^{t}}{t+1} dt$, then the value of $\int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$ is

Assertion $(A)$: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} dx}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} = \frac{\pi}{12}$
Reason $(R)$: $\int_{a}^{b} \frac{f(x) dx}{f(x)+f(a+b-x)} = \frac{b-a}{2}$

Evaluate the definite integral: $\int_0^{2a} f(x) dx$

The value of the definite integral $\int_{2}^{4} (x(3 - x)(4 + x)(6 - x)(10 - x) + \sin x) dx$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module