New Notes Blog: `int_0^(1/2) arctanx/x dx` Use a power series to approximate the value of the integral with an error of less than 0.0001.

Monday, 16 March 2015

$\displaystyle{\int_{{0}}^{{\frac{{1}}{{2}}}}}\frac{{\arctan{{x}}}}{{x}}{\left.{d}{x}\right.}$ Use a power series to approximate the value of the integral with an error of less than 0.0001.

From the Power Series table for trigonometric function, we have:

$\displaystyle{\arctan{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{{\left(-{1}\right)}}^{{n}}\frac{{{x}}^{{{2}{n}+{1}}}}{{{2}{n}+{1}}}$

$\displaystyle={x}-\frac{{{x}}^{{3}}}{{3}}+\frac{{{x}}^{{5}}}{{5}}-\frac{{{x}}^{{7}}}{{7}}+\frac{{{x}}^{{9}}}{{9}}-\ldots$

Applying it on the integral $\displaystyle{\int_{{0}}^{{\frac{{1}}{{2}}}}}\frac{{\arctan{{\left({x}\right)}}}}{{x}}{\left.{d}{x}\right.}$ where the integrand is $\displaystyle{f{{\left({x}\right)}}}=\frac{{\arctan{{\left({x}\right)}}}}{{x}}$ , we get:

$\displaystyle{\int_{{0}}^{{\frac{{1}}{{2}}}}}\frac{{\arctan{{\left({x}\right)}}}}{{x}}{\left.{d}{x}\right.}={\int_{{0}}^{{1}}}{\arctan{{\left({x}\right)}}}\cdot\frac{{1}}{{x}}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{\frac{{1}}{{2}}}}}{\sum_{{{n}={0}}}^{\infty}}{{\left(-{1}\right)}}^{{n}}\frac{{{x}}^{{{2}{n}+{1}}}}{{{2}{n}+{1}}}\cdot\frac{{1}}{{x}}{\left.{d}{x}\right.}$

...

From the Power Series table for trigonometric function, we have:

$\displaystyle{\arctan{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{{\left(-{1}\right)}}^{{n}}\frac{{{x}}^{{{2}{n}+{1}}}}{{{2}{n}+{1}}}$

$\displaystyle={x}-\frac{{{x}}^{{3}}}{{3}}+\frac{{{x}}^{{5}}}{{5}}-\frac{{{x}}^{{7}}}{{7}}+\frac{{{x}}^{{9}}}{{9}}-\ldots$

$\displaystyle{\int_{{0}}^{{\frac{{1}}{{2}}}}}\frac{{\arctan{{\left({x}\right)}}}}{{x}}{\left.{d}{x}\right.}={\int_{{0}}^{{1}}}{\arctan{{\left({x}\right)}}}\cdot\frac{{1}}{{x}}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{\frac{{1}}{{2}}}}}{\sum_{{{n}={0}}}^{\infty}}{{\left(-{1}\right)}}^{{n}}\frac{{{x}}^{{{2}{n}+{1}}}}{{{2}{n}+{1}}}\cdot\frac{{1}}{{x}}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{\frac{{1}}{{2}}}}}{\sum_{{{n}={0}}}^{\infty}}{{\left(-{1}\right)}}^{{n}}\frac{{{x}}^{{{2}{n}+{1}}}}{{{2}{n}+{1}}}\cdot{{x}}^{{-{1}}}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{\frac{{1}}{{2}}}}}{\sum_{{{n}={0}}}^{\infty}}{{\left(-{1}\right)}}^{{n}}\frac{{{x}}^{{{2}{n}+{1}-{1}}}}{{{2}{n}+{1}}}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{\frac{{1}}{{2}}}}}{\sum_{{{n}={0}}}^{\infty}}{{\left(-{1}\right)}}^{{n}}\frac{{{x}}^{{{2}{n}}}}{{{2}{n}+{1}}}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{\frac{{1}}{{2}}}}}{\left[{1}-\frac{{{x}}^{{2}}}{{3}}+\frac{{{x}}^{{4}}}{{5}}-\frac{{{x}}^{{6}}}{{7}}+\frac{{{x}}^{{8}}}{{9}}-\ldots\right]}{\left.{d}{x}\right.}$

$\displaystyle{\int_{{0}}^{{\frac{{1}}{{2}}}}}\frac{{\arctan{{\left({x}\right)}}}}{{x}}{\left.{d}{x}\right.}={\int_{{0}}^{{1}}}{\arctan{{\left({x}\right)}}}\cdot\frac{{1}}{{x}}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{\frac{{1}}{{2}}}}}\frac{{1}}{{x}}\cdot{\left[{x}-\frac{{{x}}^{{3}}}{{3}}+\frac{{{x}}^{{5}}}{{5}}-\frac{{{x}}^{{7}}}{{7}}+\frac{{{x}}^{{9}}}{{9}}-\ldots\right]}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{\frac{{1}}{{2}}}}}{\left[\frac{{x}}{{x}}-\frac{{{x}}^{{3}}}{{{3}{x}}}+\frac{{{x}}^{{5}}}{{{5}{x}}}-\frac{{{x}}^{{7}}}{{{7}{x}}}+\frac{{{x}}^{{9}}}{{{9}{x}}}-\ldots\right]}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{\frac{{1}}{{2}}}}}{\left[{1}-\frac{{{x}}^{{2}}}{{3}}+\frac{{{x}}^{{4}}}{{5}}-\frac{{{x}}^{{6}}}{{7}}+\frac{{{x}}^{{8}}}{{9}}-\ldots\right]}{\left.{d}{x}\right.}$

To determine the indefinite integral, we integrate each term using Power rule for integration: $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}$ .

$\displaystyle{\int_{{0}}^{{\frac{{1}}{{2}}}}}{\left[{1}-\frac{{{x}}^{{2}}}{{3}}+\frac{{{x}}^{{4}}}{{5}}-\frac{{{x}}^{{6}}}{{7}}+\frac{{{x}}^{{8}}}{{9}}-\ldots\right]}{\left.{d}{x}\right.}={{\left[{x}-\frac{{{x}}^{{3}}}{{{3}\cdot{3}}}+\frac{{{x}}^{{5}}}{{{5}\cdot{5}}}-\frac{{{x}}^{{7}}}{{{7}\cdot{7}}}+\frac{{{x}}^{{9}}}{{{9}\cdot{9}}}-\ldots\right]}_{{0}}^{{1}}}$

$\displaystyle={{\left[{x}-\frac{{{x}}^{{3}}}{{9}}+\frac{{{x}}^{{5}}}{{25}}-\frac{{{x}}^{{7}}}{{49}}+\frac{{{x}}^{{9}}}{{81}}-\ldots\right]}_{{0}}^{{1}}}$

Apply definite integral formula: $\displaystyle{F}{\left({x}\right)}{\mid}{{a}}^{{b}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$ .

$\displaystyle{F}{\left(\frac{{1}}{{2}}\right)}{\quad\text{or}\quad}{F}{\left({0.5}\right)}={0.5}-\frac{{{0.5}}^{{3}}}{{9}}+\frac{{{0.5}}^{{5}}}{{25}}-\frac{{{0.5}}^{{7}}}{{49}}+\frac{{{0.5}}^{{9}}}{{81}}-\ldots$

$\displaystyle=\frac{{1}}{{2}}-\frac{{1}}{{72}}+\frac{{1}}{{800}}-\frac{{1}}{{6272}}+\frac{{1}}{{41472}}-\ldots$

$\displaystyle{F}{\left({0}\right)}={0}-\frac{{{0}}^{{3}}}{{9}}+\frac{{{0}}^{{5}}}{{25}}-\frac{{{0}}^{{7}}}{{49}}+\frac{{{0}}^{{9}}}{{81}}-\ldots$

$\displaystyle={0}-{0}+{0}-{0}+{0}-\ldots$

All the terms are 0 then $\displaystyle{F}{\left({0}\right)}={0}$ .

We can stop on the 5th term $\displaystyle{\left(\frac{{1}}{{41472}}\approx{2.4113}{x}{{10}}^{{-{5}}}\right)}$ since we only need an error less than 0.0001.

$\displaystyle{F}{\left(\frac{{1}}{{2}}\right)}-{F}{\left({0}\right)}={\left[\frac{{1}}{{2}}-\frac{{1}}{{72}}+\frac{{1}}{{800}}-\frac{{1}}{{6272}}+\frac{{1}}{{41472}}\right]}-{\left[{0}\right]}$

$\displaystyle=\frac{{1}}{{2}}-\frac{{1}}{{72}}+\frac{{1}}{{800}}-\frac{{1}}{{6272}}+\frac{{1}}{{41472}}$

$\displaystyle={0.487225785}$

Then, the approximated integral value will be:

$\displaystyle{\int_{{0}}^{{\frac{{1}}{{2}}}}}\frac{{\arctan{{\left({x}\right)}}}}{{x}}{\left.{d}{x}\right.}\approx{0.4872}$

New Notes Blog

Monday, 16 March 2015

$\displaystyle{\int_{{0}}^{{\frac{{1}}{{2}}}}}\frac{{\arctan{{x}}}}{{x}}{\left.{d}{x}\right.}$ Use a power series to approximate the value of the integral with an error of less than 0.0001.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Monday, 16 March 2015

Use a power series to approximate the value of the integral with an error of less than 0.0001.

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Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{\int_{{0}}^{{\frac{{1}}{{2}}}}}\frac{{\arctan{{x}}}}{{x}}{\left.{d}{x}\right.}$ Use a power series to approximate the value of the integral with an error of less than 0.0001.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?