The value of the following definite integral is _______ (round off to three decimal places)

                                                                

The correct answer is 2.097.

To find the value of the given definite integral, we need to evaluate:
${\int }_1^e\left(x\ln x\right)dx$
as visible in the provided image.

Step 1: Find the indefinite integral using integration by parts
The formula for integration by parts is:
$\int udv=uv-\int vdu$
Let us choose:
$u=\ln x$
and
$dv=xdx$

Differentiating $u$ and integrating $dv$, we get:
$du=\frac{1}{x}dx$
and
$v=\frac{x^2}{2}$

Substituting these into the integration by parts formula:
$\int x\ln xdx=\left(\ln x\right)\left(\frac{x^2}{2}\right)-\int \frac{x^2}{2}·\frac{1}{x}dx$
Simplifying the integrand of the remaining integral:
$\int x\ln xdx=\frac{x^2}{2}\ln x-\frac{1}{2}\int xdx$
Performing the final integration:
$\int x\ln xdx=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$

Step 2: Apply the limits of integration from 1 to e
Now we evaluate the expression at the upper limit $x=e$ and the lower limit $x=1$:
${[\frac{x^2}{2}\ln x-\frac{x^2}{4}]}_1^e$

Evaluating at the upper limit $x=e$:
$(\frac{e^2}{2}\ln e-\frac{e^2}{4})=\frac{e^2}{2}\left(1\right)-\frac{e^2}{4}=\frac{e^2}{4}$

Evaluating at the lower limit $x=1$:
$(\frac{1^2}{2}\ln 1-\frac{1^2}{4})=0-\frac{1}{4}=-\frac{1}{4}$

Subtracting the lower limit value from the upper limit value:
$\frac{e^2}{4}-(-\frac{1}{4})=\frac{e^2+1}{4}$

Step 3: Calculate the numerical value
Using the approximation $e\approx 2.71828$:
$e^2\approx 7.38906$
Substitute this into the fraction:
$\frac{7.38906+1}{4}=\frac{8.38906}{4}\approx 2.09726$

Rounding off to three decimal places, we get:
2.097