Let f:[0,2] rightarrow R be a function which | vedclass

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

Question

(B) Given $F(x) = \int_0^{x^2} f(\sqrt{t}) dt$. Using the Leibniz rule for differentiation under the integral sign,we get $F'(x) = f(\sqrt{x^2}) \cdot

Vedclass · Accepted Answer

$e^4-1$

Vedclass · Answer

(B) Given $F(x) = \int_0^{x^2} f(\sqrt{t}) dt$. Using the Leibniz rule for differentiation under the integral sign,we get $F'(x) = f(\sqrt{x^2}) \cdot \frac{d}{dx}(x^2) = f(x) \cdot 2x$.Given $F'(x) = f'(x)$,we have $f'(x) = 2x f(x)$.This is a first-order linear differential equation: $\frac{f'(x)}{f(x)} = 2x$.Integrating both sides with respect to $x$,we get $\ln|f(x)| = x^2 + C$.Since $f(0) = 1$,we have $\ln(1) = 0^2 + C$,which implies $C = 0$.Thus,$\ln(f(x)) = x^2$,which gives $f(x) = e^{x^2}$.Now,$F(x) = \int_0^{x^2} e^{(\sqrt{t})^2} dt = \int_0^{x^2} e^t dt = [e^t]_0^{x^2} = e^{x^2} - e^0 = e^{x^2} - 1$.Therefore,$F(2) = e^{2^2} - 1 = e^4 - 1$.

Let $f:[0,2] \rightarrow R$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1$. Let $F(x)=\int_0^{x^2} f(\sqrt{t}) dt$ for $x \in [0,2]$. If $F'(x)=f'(x)$ for all $x \in (0,2)$,then $F(2)$ equals

Similar Questions

In a bank,the principal increases continuously at a rate of $x \%$ per year. Then the rate,$x$,if ₹$100$ doubles itself in $10$ years,is (Given $\log 2 = 0.6931$) (in $\%$)

The rate of disintegration of a radioactive element at time $t$ is proportional to its mass at that time. Then the time during which the original mass of $6 \text{ gm}$ will disintegrate into its mass of $3 \text{ gm}$ is proportional to

In a certain culture of bacteria,the rate of increase is proportional to the number present. It is found that the number doubles in $4$ hours. Then the number of times the bacteria are increased in $12$ hours is

Number of straight lines which satisfy the differential equation $\frac{dy}{dx} + x \left( \frac{dy}{dx} \right)^2 - y = 0$ is:

If a body cools from $135^{\circ} C$ to $80^{\circ} C$ at a room temperature of $25^{\circ} C$ in $60 \text{ minutes}$,then the temperature of the body after $2 \text{ hours}$ is: (in $^{\circ} C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module