Compute the definite integral:

\({\int_{{{0.2}}}^{{{2.2}}}}{x}{e}^{x}{\left.{d}{x}\right.}\)

For the integrand \(e^x x\), integrate by parts, integral \(f dg = f g\) - integral g df, where

\(\displaystyle{f}={x},{d}{g}={e}^{{{x}}}{\left.{d}{x}\right.},{d}{f}={\left.{d}{x}\right.},{g}={e}^{{{x}}}\):

\(= e^{x} x \mid_{0.2}^{2.2}-\int_{{0.2}}^{{2.2}}{e}^{x}{\left.{d}{x}\right.}\)

Evaluate the antiderivative at the limits and subtract.

\(e^{x} x \mid_{0.2}^{2.2}={e}^{2.2}{2.2}-{e}^{0.2}{0.2}={19.6107}\):

\(={19.6107}-{\int_{{0.2}}^{{2.2}}}{e}^{x}{\left.{d}{x}\right.}\)

Apply the fundamental theorem of calculus.

The antiderivative of \(\displaystyle{e}^{{{x}}}\) is \(\displaystyle{e}^{{{x}}}\):

\(= 19.6107 + (-e^{x}) \mid _{0.2}^{2.2}\)

Evaluate the antiderivative at the limits and subtract.

\((-e^{x}) \mid {0.2}^{2.2}={\left(-{e}^{2.2}\right)}-{\left(-{e}^{0.2}\right)}=-{7.80361}\):

Answer:

\(= 11.8071\)