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

For a real number $x$,let $[x]$ denote the greatest integer less than or equal to $x$ and $\{x\} = x - [x]$. Let $n$ be a positive integer. Then,$\int_0^n \cos(2 \pi [x] \{x\}) dx$ is equal to

Let $[t]$ denote the greatest integer $\leq t$. Then the value of $8 \cdot \int \limits_{-\frac{1}{2}}^{1}([2 x]+|x|) \,d x$ is .... .

$\int_{-1}^{2} {|x|\,dx} =$ (in $/2$)

If $\int_{n}^{n+1} g(x) dx = n^2, \forall n \in Z$,then the value of $\int_{-3}^3 g(x) dx$ is

$\int_0^{b - c} f''(x + a) \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module