The Notes: `int_0^3 x/sqrt(36 - x^2) dx` Evaluate the integral

Sunday, November 15, 2015

$\displaystyle{\int_{{0}}^{{3}}}\frac{{x}}{\sqrt{{{36}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$ Evaluate the integral

You need to perform the following substitution, such that $\displaystyle{36}-{{x}}^{{2}}={t}\Rightarrow-{2}{x}{\left.{d}{x}\right.}={\left.{d}{t}\right.}\Rightarrow{x}{\left.{d}{x}\right.}=-\frac{{{\left.{d}{t}\right.}}}{{2}}$

Replacing the variable yields:

$\displaystyle{\int_{{0}}^{{3}}}\frac{{{x}{\left.{d}{x}\right.}}}{{\sqrt{{{36}-{{x}}^{{2}}}}}}=-{\int_{{{t}_{{1}}}}^{{{t}_{{2}}}}}\frac{{{\left.{d}{t}\right.}}}{{{2}\sqrt{{t}}}}=-\sqrt{{t}}{{\mid}_{{{t}_{{1}}}}^{{{t}_{{2}}}}}$

Replacing back the variable yields:

$\displaystyle{\int_{{0}}^{{3}}}\frac{{{x}{\left.{d}{x}\right.}}}{{\sqrt{{{36}-{{x}}^{{2}}}}}}=-\sqrt{{{36}-{{x}}^{{2}}}}{{\mid}_{{0}}^{{3}}}$

$\displaystyle{\int_{{0}}^{{3}}}\frac{{{x}{\left.{d}{x}\right.}}}{{\sqrt{{{36}-{{x}}^{{2}}}}}}=-\sqrt{{{36}-{9}}}+\sqrt{{{36}-{0}}}$

$\displaystyle{\int_{{0}}^{{3}}}\frac{{{x}{\left.{d}{x}\right.}}}{{\sqrt{{{36}-{{x}}^{{2}}}}}}={6}-\sqrt{{27}}$

$\displaystyle{\int_{{0}}^{{3}}}\frac{{{x}{\left.{d}{x}\right.}}}{{\sqrt{{{36}-{{x}}^{{2}}}}}}={6}-{3}\sqrt{{3}}$

$\displaystyle{\int_{{0}}^{{3}}}\frac{{{x}{\left.{d}{x}\right.}}}{{\sqrt{{{36}-{{x}}^{{2}}}}}}={3}{\left({2}-\sqrt{{3}}\right)}$

Hence, evaluating the definite integral, yields $\displaystyle{\int_{{0}}^{{3}}}\frac{{{x}{\left.{d}{x}\right.}}}{{\sqrt{{{36}-{{x}}^{{2}}}}}}={3}{\left({2}-\sqrt{{3}}\right)}.$

The Notes

Sunday, November 15, 2015

$\displaystyle{\int_{{0}}^{{3}}}\frac{{x}}{\sqrt{{{36}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$ Evaluate the integral

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