Evaluate the definite integrals _0^1 ( x e^x + x 4 )dx - Vidyadip

Question

Evaluate the definite integrals $\int\limits_0^1 {\left( {x{e^x} + \sin \frac{{\pi x}}{4}} \right)dx} $

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Answer

$\int\limits_0^1 {\left( {x{e^x} + \sin \frac{{\pi x}}{4}} \right)dx} =$ $\int\limits_0^1 {x{e^x}dx + \int\limits_0^1 {\sin \frac{{\pi x}}{4}dx} } $
[Applying Product Rule on first definite integral]
$= \left( {x{e^x}} \right)_0^1 - \int\limits_0^1 {1.{e^x}dx - \frac{{\left( {\cos \frac{{\pi x}}{4}} \right)_0^1}}{{{ }{} {\text{ }}\frac{\pi }{4}}}} $
$ = {e^1} - 0 - \int\limits_0^1 {{e^x}dx - \frac{4}{\pi }\left[ {\cos \frac{\pi }{4} - \cos {0^o}} \right]} $
$= e - \left( {{e^x}} \right)_0^1 - \frac{4}{\pi }\left( {\frac{1}{{\sqrt 2 }} - 1} \right)$
$ = e - \left( {e - {e^0}} \right) - \frac{4}{{\pi \sqrt 2 }} + \frac{4}{\pi }$
$ = e - e + 1 - \frac{{2.2}}{{\pi \sqrt 2 }} + \frac{4}{\pi }$
$ = 1 + \frac{4}{\pi } - \frac{{2\sqrt 2 }}{\pi }$

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