If intlimits_0^x {fleft( t right)} dt = {x^2} | vedclass

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

Question

(A) Given the equation: $\int\limits_0^x {f\left( t \right)} dt = {x^2} + \int\limits_x^1 {{t^2}f\left( t \right)dt}$.Differentiating both sides with

Vedclass · Accepted Answer

$\frac{24}{25}$

Vedclass · Answer

(A) Given the equation: $\int\limits_0^x {f\left( t ight)} dt = {x^2} + \int\limits_x^1 {{t^2}f\left( t ight)dt}$.Differentiating both sides with respect to $x$ using the Leibniz integral rule:$\frac{d}{dx} \int\limits_0^x {f\left( t ight)} dt = \frac{d}{dx} ({x^2}) + \frac{d}{dx} \int\limits_x^1 {{t^2}f\left( t ight)dt}$.$f(x) = 2x - x^2 f(x)$.Rearranging the terms to solve for $f(x)$:$f(x) + x^2 f(x) = 2x$.$f(x)(1 + x^2) = 2x$.$f(x) = \frac{2x}{1 + x^2}$.Now,find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{(1 + x^2)(2) - (2x)(2x)}{(1 + x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2}$.Substitute $x = 1/2$:$f'(1/2) = \frac{2(1 - (1/2)^2)}{(1 + (1/2)^2)^2} = \frac{2(1 - 1/4)}{(1 + 1/4)^2} = \frac{2(3/4)}{(5/4)^2} = \frac{3/2}{25/16} = \frac{3}{2} 	imes \frac{16}{25} = \frac{24}{25}$.

If $\int\limits_0^x {f\left( t \right)} dt = {x^2} + \int\limits_x^1 {{t^2}f\left( t \right)dt} $,then $f'(1/2)$ is

Similar Questions

Let $g(x) = \int_{x}^{2x} \frac{f(t)}{t} dt$ where $x > 0$ and $f$ is a continuous function such that $f(2x) = f(x)$. Then:

The value of the integral $\int_0^{\pi / 2} \sin^5 x \, dx$ is

$\int_0^{\pi/6} \cos^4 3\theta \cdot \sin^2 6\theta \, d\theta$ is equal to

If $\int f(x) dx = F(x) + C$,then $\frac{d}{dt} \int_{g(t)}^{h(t)} f(x) dx =$

Let $f(x) = \int\limits_0^{x^2} {(t - 1)(t - 4)(t - 9)} dt$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module