Blog: `int_0^1 r^3 / sqrt(4 + r^2) dr` Evaluate the integral

Friday, March 11, 2011

$\displaystyle{\int_{{0}}^{{1}}}\frac{{{r}}^{{3}}}{\sqrt{{{4}+{{r}}^{{2}}}}}{d}{r}$ Evaluate the integral

$\displaystyle{\int_{{0}}^{{1}}}\frac{{{r}}^{{3}}}{\sqrt{{{4}+{{r}}^{{2}}}}}{d}{r}$

Let's first evaluate the indefinite integral using the method of substitution,

Substitute $\displaystyle{x}={4}+{{r}}^{{2}},\Rightarrow{{r}}^{{2}}={x}-{4}$

$\displaystyle\Rightarrow{\left.{d}{x}\right.}={2}{r}{d}{r}$

$\displaystyle\int\frac{{{r}}^{{3}}}{\sqrt{{{4}+{{r}}^{{2}}}}}{d}{r}=\int\frac{{{x}-{4}}}{{{2}\sqrt{{{x}}}}}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{1}}{{2}}\int{\left(\frac{{x}}{\sqrt{{{x}}}}-\frac{{4}}{\sqrt{{{x}}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{1}}{{2}}\int{\left(\sqrt{{{x}}}-\frac{{4}}{\sqrt{{{x}}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{1}}{{2}}{\left({\left(\frac{{{x}}^{{\frac{{1}}{{2}}+{1}}}}{{\frac{{1}}{{2}}+{1}}}\right)}-{4}{\left(\frac{{{x}}^{{-\frac{{1}}{{2}}+{1}}}}{{-\frac{{1}}{{2}}+{1}}}\right)}\right)}$

$\displaystyle=\frac{{1}}{{2}}{\left({\left(\frac{{{x}}^{{\frac{{3}}{{2}}}}}{{\frac{{3}}{{2}}}}\right)}-{4}{\left(\frac{{{x}}^{{\frac{{1}}{{2}}}}}{{\frac{{1}}{{2}}}}\right)}\right)}$

$\displaystyle=\frac{{{x}}^{{\frac{{3}}{{2}}}}}{{3}}-{4}{{x}}^{{\frac{{1}}{{2}}}}$

substitute back $\displaystyle{x}={{r}}^{{2}}+{4}$ and add constant C to the solution,

$\displaystyle=\frac{{{\left({{r}}^{{2}}+{4}\right)}}^{{\frac{{3}}{{2}}}}}{{3}}-{4}{{\left({{r}}^{{2}}+{4}\right)}}^{{\frac{{1}}{{2}}}}+{C}$

Now let's evaluate the definite integral,

$\displaystyle{\int_{{0}}^{{1}}}\frac{{{r}}^{{3}}}{\sqrt{{{4}+{{r}}^{{2}}}}}{d}{r}={{\left[\frac{{{\left({{r}}^{{2}}+{4}\right)}}^{{\frac{{3}}{{2}}}}}{{3}}-{4}{{\left({{r}}^{{2}}+{4}\right)}}^{{\frac{{1}}{{2}}}}\right]}_{{0}}^{{1}}}$

$\displaystyle={\left[\frac{{{\left({{1}}^{{2}}+{4}\right)}}^{{\frac{{3}}{{2}}}}}{{3}}-{4}{{\left({{1}}^{{2}}+{4}\right)}}^{{\frac{{1}}{{2}}}}\right]}-{\left[\frac{{{\left({{0}}^{{2}}+{4}\right)}}^{{\frac{{3}}{{2}}}}}{{3}}-{4}{{\left({{0}}^{{2}}+{4}\right)}}^{{\frac{{1}}{{2}}}}\right]}$

$\displaystyle={\left[\frac{{{\left({5}\right)}}^{{\frac{{3}}{{2}}}}}{{3}}-{4}{{\left({5}\right)}}^{{\frac{{1}}{{2}}}}\right]}-{\left[\frac{{{4}}^{{\frac{{3}}{{2}}}}}{{3}}-{4}{\left({{4}}^{{\frac{{1}}{{2}}}}\right]}\right.}$

$\displaystyle={\left[\frac{{{{5}}^{{\frac{{3}}{{2}}}}-{12}{{\left({5}\right)}}^{{\frac{{1}}{{2}}}}}}{{3}}\right]}-{\left[\frac{{{\left({{2}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}}{{3}}-{4}\cdot{2}\right]}$

$\displaystyle={\left[{{5}}^{{\frac{{1}}{{2}}}}{\left(\frac{{{5}-{12}}}{{3}}\right)}\right]}-{\left[\frac{{{2}}^{{3}}}{{3}}-{8}\right]}$

$\displaystyle={\left[-\frac{{7}}{{3}}\sqrt{{{5}}}\right]}-{\left[\frac{{8}}{{3}}-{8}\right]}$

$\displaystyle={\left(-\frac{{7}}{{3}}\sqrt{{{5}}}\right)}-{\left(\frac{{{8}-{24}}}{{3}}\right)}$

$\displaystyle=-\frac{{7}}{{3}}\sqrt{{{5}}}-{\left(-\frac{{16}}{{3}}\right)}$

$\displaystyle=-\frac{{7}}{{3}}\sqrt{{{5}}}+\frac{{16}}{{3}}$

$\displaystyle=\frac{{1}}{{3}}{\left({16}-{7}\sqrt{{{5}}}\right)}$

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Friday, March 11, 2011

$\displaystyle{\int_{{0}}^{{1}}}\frac{{{r}}^{{3}}}{\sqrt{{{4}+{{r}}^{{2}}}}}{d}{r}$ Evaluate the integral

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