Let for some function y=f(x),int_0^x t f(t) d | vedclass

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

Question

(A) Given the equation: $\int_0^{x} t f(t) dt = x^2 f(x)$ for $x > 0$. Differentiating both sides with respect to $x$ using the Leibniz rule: $x f(x)

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the equation: $\int_0^{x} t f(t) dt = x^2 f(x)$ for $x > 0$. Differentiating both sides with respect to $x$ using the Leibniz rule: $x f(x) = x^2 f'(x) + 2x f(x)$. Rearranging the terms: $x^2 f'(x) = x f(x) - 2x f(x) = -x f(x)$. Since $x > 0$,we can divide by $x^2$: $\frac{f'(x)}{f(x)} = -\frac{1}{x}$. Integrating both sides with respect to $x$: $\int \frac{f'(x)}{f(x)} dx = -\int \frac{1}{x} dx$. $\ln|f(x)| = -\ln|x| + C = \ln|\frac{k}{x}|$,where $C = \ln|k|$. Thus,$f(x) = \frac{k}{x}$. Given $f(2) = 3$,we have $3 = \frac{k}{2}$,which implies $k = 6$. Therefore,$f(x) = \frac{6}{x}$. Finally,$f(6) = \frac{6}{6} = 1$.

Let for some function $y=f(x)$,$\int_0^x t f(t) d t=x^2 f(x)$,$x > 0$ and $f(2)=3$. Then $f(6)$ is equal to :

Similar Questions

The correct evaluation of $\int_0^\pi {\left| {\,{{\sin }^4}x\,} \right|\,dx} $ is

$\int_0^\pi x \cdot \sin^5 x \cdot \cos^6 x \, dx =$

$\int_0^1 x^{3/2} \sqrt{1-x} \, dx$ is equal to

Let $f(x) = \int_{1}^{x} \sqrt{2 - t^2} dt$. Then the real roots of the equation $x^2 - f'(x) = 0$ are

The value of $\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \cos \left(t^{2}\right) d t}{x \sin x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module