Blog: `int s 2^s ds` Evaluate the integral

You need to use the integration by parts such that:

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

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

$\displaystyle{d}{g{=}}{{2}}^{{s}}\Rightarrow{g{=}}\frac{{{{2}}^{{s}}}}{{{\ln{{2}}}}}$

$\displaystyle\int{s}\cdot{{2}}^{{s}}{d}{s}={s}\cdot\frac{{{{2}}^{{s}}}}{{{\ln{{2}}}}}-\int\frac{{{{2}}^{{s}}}}{{{\ln{{2}}}}}{d}{s}$

$\displaystyle\int{s}\cdot{{2}}^{{s}}{d}{s}={s}\cdot\frac{{{{2}}^{{s}}}}{{{\ln{{2}}}}}-\frac{{{{2}}^{{s}}}}{{{{\ln}}^{{2}}{2}}}{\)}+{c}$

Hence, evaluating the integral, using integration by parts, yields $\displaystyle\int{s}\cdot{{2}}^{{s}}{d}{s}=\frac{{{{2}}^{{s}}}}{{{\ln{{2}}}}}{\left({s}-\frac{{1}}{{{\ln{{2}}}}}\right)}+{c}$

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Friday, January 14, 2011

$\displaystyle\int{s}{{2}}^{{s}}{d}{s}$ Evaluate the integral

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