int_{ - 3}^3 {frac{{{x^2}sin 2x}}{{{x^2} + 1} | vedclass

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

Question

(A) Let $f(x) = \frac{{{x^2}\sin 2x}}{{{x^2} + 1}}$.Check if the function is odd or even by evaluating $f(-x)$:$f(-x) = \frac{{{{( - x)}^2}\sin (2( -

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $f(x) = \frac{{{x^2}\sin 2x}}{{{x^2} + 1}}$.Check if the function is odd or even by evaluating $f(-x)$:$f(-x) = \frac{{{{( - x)}^2}\sin (2( - x))}}{{{{( - x)}^2} + 1}} = \frac{{{x^2}\sin ( - 2x)}}{{{x^2} + 1}} = \frac{{ - {x^2}\sin 2x}}{{{x^2} + 1}} = -f(x)$.Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{ - a}^a {f(x)\,dx = 0}$.Therefore,$\int_{ - 3}^3 {\frac{{{x^2}\sin 2x}}{{{x^2} + 1}}\,dx = 0}$.

$\int_{ - 3}^3 {\frac{{{x^2}\sin 2x}}{{{x^2} + 1}}\,dx = } $

Similar Questions

If $\int_{-1}^4 f(x) dx = 4$ and $\int_2^4 (3 - f(x)) dx = 7$,then $\int_{-1}^2 f(x) dx = $

By using the properties of definite integrals,evaluate the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{7} x \, dx$.

$\int_{0}^{1} (1 + |\sin x|)(ax^2 + bx + c) dx = \int_{0}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx$. Then,the location of the roots of $ax^2 + bx + c = 0$ is:

If $f(x) = f(2 - x)$,then $\int_{0.5}^{1.5} xf(x) dx$ equals

$ \int_{-3}^{3} \cot^{-1} x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module