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

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

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hesgidiauE

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.
$$(-e^{x}) \mid {0.2}^{2.2}={\left(-{e}^{2.2}\right)}-{\left(-{e}^{0.2}\right)}=-{7.80361}$$:
$$= 11.8071$$