int_0^1 {{e^{2ln x}}dx} = | vedclass

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

Question

(C) We are given the integral $I = \int_0^1 {{e^{2\ln x}}dx}$.Using the logarithmic property $a \ln b = \ln(b^a)$,we can rewrite the exponent:$2\ln x

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(C) We are given the integral $I = \int_0^1 {{e^{2\ln x}}dx}$.Using the logarithmic property $a \ln b = \ln(b^a)$,we can rewrite the exponent:$2\ln x = \ln(x^2)$.Substituting this into the integral,we get:$I = \int_0^1 {{e^{\ln(x^2)}}dx}$.Since $e^{\ln(f(x))} = f(x)$,the integral simplifies to:$I = \int_0^1 {{x^2}dx}$.Now,we evaluate the definite integral:$I = \left[ \frac{x^3}{3} \right]_0^1$.Applying the limits:$I = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} - 0 = \frac{1}{3}$.

$\int_0^1 {{e^{2\ln x}}dx} = $

Similar Questions

If $[x]$ denotes the greatest integer less than or equal to $x,$ then the value of the integral $\int_0^2 {{x^2}[x]\,dx} $ equals (in $/3$)

$\int_{-1}^2 |x| \, dx =$

Evaluate the definite integral $\int_{2}^{3} \frac{1}{x} d x$.

Let $[\cdot]$ denote the greatest integer function. Then the value of $\int_0^3 \left( \frac{e^x + e^{-x}}{[x]!} \right) dx$ is :

$\int_{a-c}^{b-c} f(x+c) \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module