Blog: `int y/(e^(2y)) dy` Evaluate the integral

Saturday, June 12, 2010

$\displaystyle\int\frac{{y}}{{{{e}}^{{{2}{y}}}}}{\left.{d}{y}\right.}$ Evaluate the integral

You need to use the substitution $\displaystyle-{2}{y}={u}$ , such that:

$\displaystyle-{2}{y}={u}\Rightarrow-{2}{\left.{d}{y}\right.}={d}{u}\Rightarrow{\left.{d}{y}\right.}=-\frac{{{d}{u}}}{{{2}}}$

Replacing the variable, yields:

$\displaystyle\int{y}\cdot{{e}}^{{-{2}{y}}}{\left.{d}{y}\right.}={\left(\frac{{1}}{{4}}\right)}\int{u}\cdot{{e}}^{{u}}{d}{u}$

You need to use the integration by parts such that:

$\displaystyle\int{f{{d}}}{g{=}}{f{{g{-}}}}\int{g{{d}}}{f{}}$

$\displaystyle{f{=}}{u}\Rightarrow{d}{f{=}}{d}{u}$

$\displaystyle{d}{g{=}}{{e}}^{{u}}\Rightarrow{g{=}}{{e}}^{{u}}$

$\displaystyle{\left(\frac{{1}}{{4}}\right)}\int{u}\cdot{{e}}^{{u}}{d}{u}={\left(\frac{{1}}{{4}}\right)}{\left({u}\cdot{{e}}^{{u}}-\int{{e}}^{{u}}{d}{u}\right)}$

$\displaystyle{\left(\frac{{1}}{{4}}\right)}\int{u}\cdot{{e}}^{{u}}{d}{u}={\left(\frac{{1}}{{4}}\right)}{u}\cdot{{e}}^{{u}}-{\left(\frac{{1}}{{4}}\right)}{{e}}^{{u}}+{c}$

Replacing back the variable, yields:

$\displaystyle\int{y}\cdot{{e}}^{{-{2}{y}}}{\left.{d}{y}\right.}={\left(\frac{{1}}{{4}}\right)}{\left({\left(-{2}{y}\right)}\cdot{{e}}^{{-{2}{y}}}-{{e}}^{{-{2}{y}}}\right)}+{c}$

Hence, evaluating the integral, using substitution, then integration by parts, yields $\displaystyle\int{y}\cdot{{e}}^{{-{2}{y}}}{\left.{d}{y}\right.}={\left(\frac{{{{e}}^{{-{2}{y}}}}}{{4}}\right)}{\left(-{2}{y}-{1}\right)}+{c}$

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Saturday, June 12, 2010

$\displaystyle\int\frac{{y}}{{{{e}}^{{{2}{y}}}}}{\left.{d}{y}\right.}$ Evaluate the integral

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