Question

The value of the following definite integral is (round off to three decimal places).
${\int }_1^e\left(xlnx\right)dx$

• A. 2.09

Answer 1

The correct option is A 2.09
${\int }_1^{e}xlnxdx$
$=lnx.\frac{x^2}{2}-\int [\frac{1}{x}\times \frac{x^2}{2}]dx$
${\begin{array}{c}=\frac{x^{2}lnx}{2}-\frac{x^2}{4}\end{array}|}_1^e$
$=(\frac{e^2}{2}-\frac{e^2}{4})+\frac{1}{4}$
$=\frac{e^2+1}{4}=2.907$