int_{-1}^1 (a x^3 + b x) dx = 0 for | vedclass

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

Question

(A) Let $f(x) = a x^3 + b x$. Since $f(-x) = a(-x)^3 + b(-x) = -(a x^3 + b x) = -f(x)$, the function $f(x)$ is an odd function. According to the prope

Vedclass · Accepted Answer

any value of $a$ and $b$

Vedclass · Answer

(A) Let $f(x) = a x^3 + b x$. Since $f(-x) = a(-x)^3 + b(-x) = -(a x^3 + b x) = -f(x)$, the function $f(x)$ is an odd function. According to the property of definite integrals, if $f(x)$ is an odd function, then $\int_{-a}^a f(x) dx = 0$. Alternatively, evaluating the integral: $\int_{-1}^1 (a x^3 + b x) dx = \left[ a \frac{x^4}{4} + b \frac{x^2}{2} \right]_{-1}^1$ $= \left( \frac{a(1)^4}{4} + \frac{b(1)^2}{2} \right) - \left( \frac{a(-1)^4}{4} + \frac{b(-1)^2}{2} \right)$ $= \left( \frac{a}{4} + \frac{b}{2} \right) - \left( \frac{a}{4} + \frac{b}{2} \right) = 0$. This result holds true for any real values of $a$ and $b$.

$\int_{-1}^1 (a x^3 + b x) dx = 0$ for

Similar Questions

The value of $\sum_{n=1}^{10} \int_{-2n-1}^{-2n} \sin^{27} x \, dx + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx$ is equal to

$\int_{0}^{\frac{\pi}{2}} \frac{1-\cot x}{\operatorname{cosec} x+\cos x} d x=$

$ \int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1-\sin x \cos x} d x = $

The value of the integral $\int_{-2}^{2} \frac{|x^{3}+x|}{e^{x|x|}+1} dx$ is equal to

$\int_{0}^{2a} f(x) \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module