Question

Evaluate the integral ${\int }_1^3\left|\right(2-x\left){\log }_{e}x\right|dx$

• A. $\frac{3}{2}{\log }_{e}3+\frac{1}{2}$
• B. ${\log }_e\frac{16}{3\sqrt{2}}-\frac{1}{2}$
• C. $-\frac{3}{2}{\log }_{e}3-\frac{1}{2}$
• D. none of these

Answer 1

The correct option is B ${\log }_e\frac{16}{3\sqrt{2}}-\frac{1}{2}$

${\int }_1^3\left|(2-x){\log }_{e}x\right|dx$

$={\int }_1^2\left((2-x){\log }_{e}x\right)dx+{\int }_2^3(x-2){\log }_{e}xdx$
$={\left[{\log }_{e}x(2x-\frac{x^2}{2})\right]}_1^2-{\int }_1^2(2x-\frac{x^2}{2})\frac{1}{x}dx$
$+{\left[{\log }_{e}x(\frac{x^2}{2}-2x)\right]}_2^3-{\int }_2^3(\frac{x^2}{2}-2x)\frac{1}{x}dx$
$=\log \frac{16}{3\sqrt{2}}-\frac{1}{2}$