Blog: `int (1 - 2x)^9 dx` Evaluate the indefinite integral.

Thursday, December 19, 2013

$\displaystyle\int{{\left({1}-{2}{x}\right)}}^{{9}}{\left.{d}{x}\right.}$ Evaluate the indefinite integral.

You need to evaluate the indefinite integral by performing the substitution 1 - 2x = t, such that:

$\displaystyle{1}-{2}{x}={t}\Rightarrow-{2}{\left.{d}{x}\right.}={\left.{d}{t}\right.}\Rightarrow{\left.{d}{x}\right.}=-\frac{{{\left.{d}{t}\right.}}}{{2}}$

$\displaystyle\int{{\left({1}-{2}{x}\right)}}^{{9}}{\left.{d}{x}\right.}=-\frac{{1}}{{2}}\int{{t}}^{{9}}{\left.{d}{t}\right.}$

You need to use the following formula of integration, such that:

$\displaystyle\int{{t}}^{{n}}{\left.{d}{t}\right.}=\frac{{{{t}}^{{{n}+{1}}}}}{{{n}+{1}}}$

$\displaystyle-\frac{{1}}{{2}}\int{{t}}^{{9}}{\left.{d}{t}\right.}=-{\left(\frac{{1}}{{2}}\right)}\frac{{{{t}}^{{{9}+{1}}}}}{{{9}+{1}}}+{c}$

Replacing back 1 - 2x for t yields:

$\displaystyle\int{{\left({1}-{2}{x}\right)}}^{{9}}{\left.{d}{x}\right.}=-{\left(\frac{{1}}{{2}}\right)}\frac{{{{\left({1}-{2}{x}\right)}}^{{10}}}}{{10}}+{c}$

Hence, evaluating the indefinite integral yields $\displaystyle\int{{\left({1}-{2}{x}\right)}}^{{9}}{\left.{d}{x}\right.}=-\frac{{{{\left({1}-{2}{x}\right)}}^{{10}}}}{{20}}+{c}.$

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Thursday, December 19, 2013

$\displaystyle\int{{\left({1}-{2}{x}\right)}}^{{9}}{\left.{d}{x}\right.}$ Evaluate the indefinite integral.

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