If int_0^x {f(t),dt} = x + int_x^1 {t,f(t),dt | vedclass

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

Question

(A) Given the equation: $\int_0^x {f(t)\,dt} = x + \int_x^1 {t\,f(t)\,dt}$ Rewrite the integral $\int_x^1 {t\,f(t)\,dt}$ as $-\int_1^x {t\,f(t)\,dt}$:

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(A) Given the equation: $\int_0^x {f(t)\,dt} = x + \int_x^1 {t\,f(t)\,dt}$ Rewrite the integral $\int_x^1 {t\,f(t)\,dt}$ as $-\int_1^x {t\,f(t)\,dt}$: $\int_0^x {f(t)\,dt} = x - \int_1^x {t\,f(t)\,dt}$ Differentiating both sides with respect to $x$ using the Leibniz integral rule: $\frac{d}{dx} \left( \int_0^x {f(t)\,dt} \right) = \frac{d}{dx} (x) - \frac{d}{dx} \left( \int_1^x {t\,f(t)\,dt} \right)$ Applying the Fundamental Theorem of Calculus: $f(x) = 1 - xf(x)$ Rearranging the terms to solve for $f(x)$: $f(x) + xf(x) = 1$ $f(x)(1 + x) = 1$ $f(x) = \frac{1}{1 + x}$ Now,substitute $x = 1$ to find $f(1)$: $f(1) = \frac{1}{1 + 1} = \frac{1}{2}$.

If $\int_0^x {f(t)\,dt} = x + \int_x^1 {t\,f(t)\,dt,}$ then the value of $f(1)$ is

Similar Questions

The points of extremum of $\int_0^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \,dt$ are

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

$\int_{-\pi / 2}^{\pi / 2} \sin ^4 x \cos ^6 x \, dx$ is equal to

The function $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$ is an increasing function in the interval ...

The points of extremum of $\int_0^{x^2} \frac{t^2-5t+4}{2+e^t} dt$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module