int_{-pi / 4}^{pi / 4} x^3 sin ^4(x) d x= | vedclass

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

Question

(A) Let $I = \int_{-\pi / 4}^{\pi / 4} x^3 \sin^4(x) dx$. We know that for a definite integral,$\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd functio

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $I = \int_{-\pi / 4}^{\pi / 4} x^3 \sin^4(x) dx$. We know that for a definite integral,$\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function,i.e.,$f(-x) = -f(x)$. Here,$f(x) = x^3 \sin^4(x)$. Calculating $f(-x)$: $f(-x) = (-x)^3 \sin^4(-x) = -x^3 (\sin(x))^4 = -x^3 \sin^4(x) = -f(x)$. Since $f(x)$ is an odd function,the integral over the symmetric interval $[-\pi/4, \pi/4]$ is $0$. Therefore,$I = 0$.

$\int_{-\pi / 4}^{\pi / 4} x^3 \sin ^4(x) d x=$

Similar Questions

If $\int_{0}^{1} 4 \cot^{-1}(1-x+x^{2}) dx = a \tan^{-1}(2) - b \log_{e}(5)$,where $a, b \in N$,then $(2a+b)$ is equal to :

Evaluate the integral: $\int_{-1}^{1} \left[ \sqrt{1+x+x^{2}} - \sqrt{1-x+x^{2}} \right] dx$

$\int_{-1/2}^{1/2} \log \left(\frac{1+x}{1-x}\right) dx=$

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x^3+\cos x+\tan^5 x) dx$ is equal to . . . . . . .

$\int_{0}^{\pi} x f(\sin x) dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module