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

Thursday, March 8, 2012

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

Recall that first fundamental theorem of calculus indicates that $\displaystyle{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}{{\mid}_{{a}}^{{b}}}:$

$\displaystyle{f{{\left({x}\right)}}}$ as the integrand function

$\displaystyle{F}{\left({x}\right)}$ as the antiderivative of $\displaystyle{f{{\left({x}\right)}}}$

" $\displaystyle{a}$ " as the lower boundary value of $\displaystyle{x}$

" $\displaystyle{b}$ " as the upper boundary value of $\displaystyle{x}$

To evaluate the given problem: $\displaystyle{\int_{{0}}^{{{4}}}}\frac{{1}}{{{25}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$ , we may rewrite in a form of:

$\displaystyle{\int_{{0}}^{{{4}}}}\frac{{1}}{{{{5}}^{{2}}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$ .

The integral part resembles the integration formula for inverse of hyperbolic tangent function: $\displaystyle\int\frac{{1}}{{{{a}}^{{2}}-{{u}}^{{2}}}}{d}{u}={\left(\frac{{1}}{{a}}\right)}{\arctan{{h}}}{\left(\frac{{u}}{{a}}\right)}+$ C.

By comparison, it shows that $\displaystyle{{a}}^{{2}}$ corresponds to $\displaystyle{{5}}^{{2}}$ and $\displaystyle{{u}}^{{2}}$ corresponds to $\displaystyle{{x}}^{{2}}$ . Therefore, it shows that $\displaystyle{a}={5}$ and $\displaystyle{u}={x}$ .

By following the formula, the indefinite integral function will be:

$\displaystyle{\int_{{0}}^{{{4}}}}\frac{{1}}{{{{5}}^{{2}}-{{x}}^{{2}}}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{5}}\right)}{\arctan{{h}}}{\left(\frac{{x}}{{5}}\right)}{{\mid}_{{0}}^{{4}}}$

To solve for the definite integral, we may apply $\displaystyle{F}{\left({x}\right)}{{\mid}_{{a}}^{{b}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$ , we get:

$\displaystyle{\left(\frac{{1}}{{5}}\right)}{\arctan{{h}}}{\left(\frac{{x}}{{5}}\right)}{{\mid}_{{0}}^{{4}}}={\left(\frac{{1}}{{5}}\right)}{\arctan{{h}}}{\left(\frac{{4}}{{5}}\right)}-{\left(\frac{{1}}{{5}}\right)}{\arctan{{h}}}{\left(\frac{{0}}{{5}}\right)}$

$\displaystyle={\left(\frac{{1}}{{5}}\right)}{\arctan{{h}}}{\left(\frac{{4}}{{5}}\right)}-{\left(\frac{{1}}{{5}}\right)}{\arctan{{h}}}{\left({0}\right)}$

$\displaystyle={\left(\frac{{1}}{{5}}\right)}{a}{r}{c}{h}{\tan{{\left(\frac{{4}}{{5}}\right)}}}-{0}$

$\displaystyle=\frac{{1}}{{5}}{\arctan{{h}}}{\left(\frac{{4}}{{5}}\right)}$

The $\displaystyle\frac{{1}}{{5}}{\arctan{{h}}}{\left(\frac{{4}}{{5}}\right)}$ can be simplified as $\displaystyle{0.2197}$ as rounded off value.

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Thursday, March 8, 2012

$\displaystyle{\int_{{0}}^{{4}}}\frac{{1}}{{{25}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$ 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?

Thursday, March 8, 2012

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?

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

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