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

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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} = $

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