Let f: R rightarrow R be a continuous functio | vedclass

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

Question

(A) Given,$f(x) = \int_{0}^{x} f(t) \, dt$. Applying the Leibniz rule for differentiation under the integral sign,we differentiate both sides with res

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given,$f(x) = \int_{0}^{x} f(t) \, dt$. Applying the Leibniz rule for differentiation under the integral sign,we differentiate both sides with respect to $x$: $f'(x) = f(x)$. This is a first-order linear differential equation. Separating variables,we get $\frac{f'(x)}{f(x)} = 1$. Integrating both sides with respect to $x$,we get $\ln|f(x)| = x + C$,which implies $f(x) = k e^{x}$ for some constant $k$. Now,substitute $x = 0$ into the original equation: $f(0) = \int_{0}^{0} f(t) \, dt = 0$. Using $f(0) = k e^{0} = k$,we find $k = 0$. Therefore,$f(x) = 0 \cdot e^{x} = 0$ for all $x \in R$. Thus,$f(\log_{e} 5) = 0$.

Let $f: R \rightarrow R$ be a continuous function which satisfies $f(x) = \int_{0}^{x} f(t) \, dt$. Then,the value of $f(\log_{e} 5)$ is:

Similar Questions

The equation of the curve passing through $(1,2)$ and whose tangent at any point $(x, y)$ makes an angle $\tan ^{-1}(2 x+3 y)$ with the $X$-axis is .........

The solution of the differential equation $(x+1) \frac{dy}{dx} - xy = 1$,satisfying $y(0) = 1$ is

The integrating factor of the differential equation $\frac{dy}{dx} = y \tan x - y^2 \sec x$ is

Let $f:[1, \infty) \rightarrow \mathbb{R}$ be a differentiable function. If $6 \int_{1}^{x} f(t) dt = 3xf(x) + x^{3} - 4$ for all $x \ge 1$,then the value of $f(2) - f(3)$ is

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=2(y+2 \sin x-5)x-2 \cos x$ such that $y(0)=7$. Then $y(\pi)$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module