Let f(x) be a positive function,I_1 = int_{-f | vedclass

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

Question

(D) Given $I_1 = \int_{-\frac{1}{2}}^1 2x f(2x(1-2x)) dx$. Let $2x = t$,then $2dx = dt$,so $dx = \frac{1}{2} dt$. When $x = -\frac{1}{2}$,$t = -1$. Wh

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Given $I_1 = \int_{-\frac{1}{2}}^1 2x f(2x(1-2x)) dx$. Let $2x = t$,then $2dx = dt$,so $dx = \frac{1}{2} dt$. When $x = -\frac{1}{2}$,$t = -1$. When $x = 1$,$t = 2$. Substituting these,$I_1 = \int_{-1}^2 t f(t(1-t)) \frac{1}{2} dt = \frac{1}{2} \int_{-1}^2 t f(t(1-t)) dt$. Thus,$2I_1 = \int_{-1}^2 t f(t(1-t)) dt$. Using the property $\int_a^b g(t) dt = \int_a^b g(a+b-t) dt$,we have $2I_1 = \int_{-1}^2 (1-t) f((1-t)(1-(1-t))) dt = \int_{-1}^2 (1-t) f((1-t)t) dt$. Expanding this,$2I_1 = \int_{-1}^2 f(t(1-t)) dt - \int_{-1}^2 t f(t(1-t)) dt$. Substituting $I_2 = \int_{-1}^2 f(t(1-t)) dt$ and $2I_1 = \int_{-1}^2 t f(t(1-t)) dt$,we get $2I_1 = I_2 - 2I_1$. Therefore,$4I_1 = I_2$,which implies $\frac{I_2}{I_1} = 4$.

Let $f(x)$ be a positive function,$I_1 = \int_{-\frac{1}{2}}^1 2x f(2x(1-2x)) dx$,and $I_2 = \int_{-1}^2 f(x(1-x)) dx$. Then the value of $\frac{I_2}{I_1}$ is equal to:

Similar Questions

$\int_{0}^{\pi} \frac{x \, dx}{1+\cos \alpha \sin x}, (0 < \alpha < \pi)$ is equal to

$\int_0^{\pi /2} \frac{\cos x - \sin x}{1 + \sin x \cos x} \,dx = $

$\int_{1}^{6\pi}([\sec^{-1}x]+[\cot^{-1}x])dx$ is equal to (where $[.]$ denotes the greatest integer function).

If $\int_0^{2024 \pi} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x=k$,then $\left(\frac{2 k}{\pi}+1\right)=$

Let $m, n, p, q$ be four positive integers. If $\int_0^{2 \pi} \sin^m x \cos^n x \, dx = 4 \int_0^{\pi/2} \sin^m x \cos^n x \, dx$, $\int_0^{2 \pi} \sin^p x \cos^n x \, dx = 0$, $\int_0^{\pi} \sin^p x \cos^q x \, dx = 0$, $a = m + n + p$ and $b = m + n + q$, then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module