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The integral

Question:

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

  1. $\frac{1}{2}-\mathrm{e}-\frac{1}{\mathrm{e}^{2}}$

  2. $\frac{3}{2}-\frac{1}{\mathrm{e}}-\frac{1}{2 \mathrm{e}^{2}}$

  3. $-\frac{1}{2}+\frac{1}{\mathrm{e}}-\frac{1}{2 \mathrm{e}^{2}}$

  4. $\frac{3}{2}-\mathrm{e}-\frac{1}{2 \mathrm{e}^{2}}$


Correct Option: , 4

Solution:

$\int_{1}^{\mathrm{e}}\left(\frac{\mathrm{x}}{\mathrm{e}}\right)^{2 \mathrm{x}} \log _{\mathrm{e}} \mathrm{x} \cdot \mathrm{dx}-\int_{1}^{\mathrm{e}}\left(\frac{\mathrm{e}}{\mathrm{x}}\right) \log _{\mathrm{e}} x \cdot d x$

$\operatorname{Let}\left(\frac{x}{e}\right)^{2 x}=t,\left(\frac{e}{x}\right)^{x}=v$

$=\frac{1}{2} \int_{\left(\frac{1}{e}\right)^{2}}^{1} d t+\int_{\mathrm{e}}^{1} d v$

$=\frac{1}{2}\left(1-\frac{1}{\mathrm{e}^{2}}\right)+(1-\mathrm{e})=\frac{3}{2}-\frac{1}{2 \mathrm{e}^{2}}-\mathrm{e}$

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