Question

The value of the integral $\int_{0}^{1}$ $\frac{e^{5log_{e}x}-e^{4log_{e}x}}{e^{log_{e}x^3}-e^{log_{e}x^2}}dx$ is

• A. $\frac{1}{3}$
• B. $1$
• C. $-\frac{1}{3}$
• D. $-1$

Answer 1

Answer: (A)

$\frac{1}{3}$

Answer 2

Let $I =\int\limits_{0}^{1} \frac{e^{5 log_{e} x} -e^{4 log_{e} x}}{e^{log_{e} x^3}-e^{log_{e}x^2}} dx$
$= \int\limits_{0}^{1} \frac{e^{log_{e}^5}-e^{log_{e} x^4}}{e^{log_{e} x^3}-e^{log_{e} x^2}} dx$
$= \int\limits_{0}^{1} \frac{x^{5}-x^{4}}{x^{3}-x^{2}} dx \left[\because e^{log_{e} x} = X\right]$
$= \int\limits_{0}^{1} \frac{x^{4}\left(x-1\right)}{x^{2}\left(x-1\right)}dx$
$= \int\limits_{0}^{1} x^{2} dx$
$= \left[\frac{x^{3}}{3}\right]_{0}^{1} = \frac{1^{3}}{3} - \frac{0^{3}}{3}$
$\Rightarrow I = \frac{1}{3}$