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

ediculeN 2021-09-08 Answered

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

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

hesgidiauE
Answered 2021-09-09 Author has 16664 answers

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:
\(= 11.8071\)

Have a similar question?
Ask An Expert
29
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more
...