Question

The integral $\underset{1}{\overset{e}{\int }}\{{\left(\frac{x}{e}\right)}^{2x}-{\left(\frac{e}{x}\right)}^x\}{\log }_{e}xdx$ is equal to:

• A. $\frac{3}{2}-\frac{1}{e}-\frac{1}{2e^2}$
• B. $\frac{3}{2}-e-\frac{1}{2e^2}$
• C. $-\frac{1}{2}+\frac{1}{e}-\frac{1}{2e^2}$
• D. $\frac{1}{2}-e-\frac{1}{e^2}$

Answer 1

The correct option is B $\frac{3}{2}-e-\frac{1}{2e^2}$

$I=\underset{1}{\overset{e}{\int }}\{{\left(\frac{x}{e}\right)}^{2x}-{\left(\frac{e}{x}\right)}^x\}{\log }_{e}xdx$

$=\underset{1}{\overset{e}{\int }}{\left(\frac{x}{e}\right)}^{2x}{\log }_{e}xdx-\underset{1}{\overset{e}{\int }}{\left(\frac{e}{x}\right)}^x{\log }_{e}xdx$

Let ${\left(\frac{x}{e}\right)}^{2x}=t\Rightarrow 2{\left(\frac{x}{e}\right)}^{2x}\log xdx=dt$
and ${\left(\frac{e}{x}\right)}^x=u\Rightarrow -{\left(\frac{e}{x}\right)}^x\log xdx=du$

$\therefore I=\frac{1}{2}\underset{1/e^2}{\overset{1}{\int }}dt+\underset{e}{\overset{1}{\int }}du$
$=\frac{1}{2}(1-\frac{1}{e^2})+(1-e)$
$=\frac{3}{2}-e-\frac{1}{2e^2}$