Int of xe^x^2dx

Thursday, June 11, 2015

Int of xe^x^2dx

Evaluate $\displaystyle\int{x}{{e}}^{{{{x}}^{{2}}}}{\left.{d}{x}\right.}$ :

Let $\displaystyle{u}={{x}}^{{2}}$ so du=2xdx. Then:

$\displaystyle\int{x}{{e}}^{{{{x}}^{{2}}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}\int{{e}}^{{u}}{d}{u}$

$\displaystyle=\frac{{1}}{{2}}{{e}}^{{u}}+{C}$

Substituting back for u we get:

$\displaystyle\int{x}{{e}}^{{{{x}}^{{2}}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{{e}}^{{{{x}}^{{2}}}}+{C}$

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We can check by taking the derivative:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left[\frac{{1}}{{2}}{{e}}^{{{{x}}^{{2}}}}+{C}\right]}={\left(\frac{{1}}{{2}}\right)}{\left({{e}}^{{{{x}}^{{2}}}}\right)}{\left({2}{x}\right)}={x}{{e}}^{{{{x}}^{{2}}}}$ as required.

Ninenotes

Thursday, June 11, 2015

Int of xe^x^2dx

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