Let f be a positive function. Let I_1 = int_{ | vedclass

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

Question

(C) We are given $I_1 = \int_{1-k}^{k} x f\{x(1-x)\} dx$ and $I_2 = \int_{1-k}^{k} f\{x(1-x)\} dx$.Using the property $\int_{a}^{b} g(x) dx = \int_{a}

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(C) We are given $I_1 = \int_{1-k}^{k} x f\{x(1-x)\} dx$ and $I_2 = \int_{1-k}^{k} f\{x(1-x)\} dx$.Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$,we apply it to $I_1$ with $a = 1-k$ and $b = k$,so $a+b = 1$.$I_1 = \int_{1-k}^{k} (1-x) f\{(1-x)(1-(1-x))\} dx$$I_1 = \int_{1-k}^{k} (1-x) f\{(1-x)x\} dx$$I_1 = \int_{1-k}^{k} f\{x(1-x)\} dx - \int_{1-k}^{k} x f\{x(1-x)\} dx$$I_1 = I_2 - I_1$$2I_1 = I_2$Therefore,$\frac{I_1}{I_2} = \frac{1}{2}$.

Let $f$ be a positive function. Let $I_1 = \int_{1-k}^{k} x f\{x(1-x)\} dx$ and $I_2 = \int_{1-k}^{k} f\{x(1-x)\} dx$,where $2k-1 > 0$. Then $I_1/I_2$ is

Similar Questions

Which of the following organs is known as the 'blood bank' of the human body?

Clamp connection is found in

Which of the following coenzymes is a derivative of pantothenic acid (Vitamin $B$ complex)?

Two animals which are the members of the same order must also be the members of the same

$540\, g$ of ice at $0\,^{\circ}C$ is mixed with $540\, g$ of water at $80\,^{\circ}C$. The final temperature of the mixture is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module