`int_0^4 1/sqrt(25-x^2) dx` Evaluate the integral

Tuesday, April 15, 2014

$\displaystyle{\int_{{0}}^{{4}}}\frac{{1}}{\sqrt{{{25}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$ Evaluate the integral

The function under the integral is similar to the derivative of $\displaystyle{\arcsin{{\left({x}\right)}}},$ $\displaystyle\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}},$ and may be reduced to it.

Namely, make the substitution $\displaystyle{x}={5}{u},$ then $\displaystyle{\left.{d}{x}\right.}={5}{d}{u}.$ Also this means $\displaystyle{u}=\frac{{x}}{{5}}$ and therefore the limits of integration for $\displaystyle{u}$ are from $\displaystyle\frac{{0}}{{5}}={0}$ to $\displaystyle\frac{{4}}{{5}}$ (substitute the limits for $\displaystyle{x}$ to the formula $\displaystyle{u}=\frac{{x}}{{5}}$ ).

The integral becomes

$\displaystyle{\int_{{0}}^{{\frac{{4}}{{5}}}}}\frac{{{5}{d}{u}}}{\sqrt{{{25}-{25}{{u}}^{{2}}}}}={\int_{{0}}^{{\frac{{4}}{{5}}}}}\frac{{{d}{u}}}{\sqrt{{{1}-{{u}}^{{2}}}}}=$

$\displaystyle={\arcsin{{\left(\frac{{4}}{{5}}\right)}}}-{\arcsin{{\left({0}\right)}}}={\arcsin{{\left(\frac{{4}}{{5}}\right)}}}\approx{0.93}.$

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Tuesday, April 15, 2014

$\displaystyle{\int_{{0}}^{{4}}}\frac{{1}}{\sqrt{{{25}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$ Evaluate the integral

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Tuesday, April 15, 2014

Evaluate the integral

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Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{\int_{{0}}^{{4}}}\frac{{1}}{\sqrt{{{25}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$ Evaluate the integral

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?