For f(x) = x^4 + |x|,let I_1 = int_{0}^{pi} f | vedclass

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

Question

(C) Given $f(x) = x^4 + |x|$. Since $f(-x) = (-x)^4 + |-x| = x^4 + |x| = f(x)$,$f(x)$ is an even function.$I_1 = \int_{0}^{\pi} f(\cos x) dx$.Using th

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given $f(x) = x^4 + |x|$. Since $f(-x) = (-x)^4 + |-x| = x^4 + |x| = f(x)$,$f(x)$ is an even function.$I_1 = \int_{0}^{\pi} f(\cos x) dx$.Using the property $\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a-x) dx$,we get:$I_1 = \int_{0}^{\pi} f(\cos(\pi - x)) dx = \int_{0}^{\pi} f(-\cos x) dx$.Since $f(x)$ is an even function,$f(-\cos x) = f(\cos x)$.Thus,$I_1 = \int_{0}^{\pi} f(\cos x) dx = 2 \int_{0}^{\frac{\pi}{2}} f(\cos x) dx$.Now,consider $I_2 = \int_{0}^{\frac{\pi}{2}} f(\sin x) dx$. Using the property $\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a-x) dx$:$I_2 = \int_{0}^{\frac{\pi}{2}} f(\sin(\frac{\pi}{2} - x)) dx = \int_{0}^{\frac{\pi}{2}} f(\cos x) dx$.Therefore,$I_1 = 2 I_2$,which implies $\frac{I_1}{I_2} = 2$.

For $f(x) = x^4 + |x|$,let $I_1 = \int_{0}^{\pi} f(\cos x) dx$ and $I_2 = \int_{0}^{\frac{\pi}{2}} f(\sin x) dx$. Then $\frac{I_1}{I_2}$ is equal to

Similar Questions

The integral $\int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x$ is equal to :

The value of $\int_{e^{-1}}^{e^2} \left| \frac{\log_e x}{x} \right| dx$ is

$\int_{\pi / 11}^{9 \pi / 22} \frac{d x}{1+\sqrt{\tan x}} = $

For $n > 0,$ $\int_0^{2\pi } {\frac{{x{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}\,dx = } $

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module