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

Saturday, 26 September 2015

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

From the table of power series, we have:

$\displaystyle{{e}}^{{x}}={\sum_{{{n}={0}}}^{\infty}}\frac{{{x}}^{{n}}}{{n}}!$

$\displaystyle={1}+{x}+\frac{{{x}}^{{2}}}{{{2}!}}+\frac{{{x}}^{{3}}}{{{3}!}}+\frac{{{x}}^{{4}}}{{{4}!}}+\frac{{{x}}^{{5}}}{{{5}!}}+$ ...

To apply this on the given integral $\displaystyle{\int_{{0}}^{{1}}}{{e}}^{{-{{x}}^{{2}}}}{\left.{d}{x}\right.}$ ,

we replace the " $\displaystyle{x}$ " with " $\displaystyle-{{x}}^{{2}}$ ".

$\displaystyle{{e}}^{{-{{x}}^{{2}}}}={\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{{x}}^{{2}}\right)}}^{{n}}}{{{n}!}}$

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

$\displaystyle=\frac{{1}}{{{0}!}}-\frac{{{x}}^{{2}}}{{{1}!}}+\frac{{{x}}^{{4}}}{{{2}!}}-\frac{{{x}}^{{6}}}{{{3}!}}+\frac{{{x}}^{{8}}}{{4}}!-\frac{{{x}}^{{{10}}}}{{{5}!}}+\frac{{{x}}^{{{12}}}}{{{6}!}}$ -...

$\displaystyle={1}-{{x}}^{{2}}+\frac{{{x}}^{{4}}}{{2}}-\frac{{{x}}^{{6}}}{{6}}+\frac{{{x}}^{{8}}}{{24}}-\frac{{{x}}^{{{10}}}}{{120}}+\frac{{{x}}^{{{12}}}}{{{6}!}}-$ ...

The integral becomes:

$\displaystyle{\int_{{0}}^{{1}}}\ldots$

From the table of power series, we have:

$\displaystyle{{e}}^{{x}}={\sum_{{{n}={0}}}^{\infty}}\frac{{{x}}^{{n}}}{{n}}!$

$\displaystyle={1}+{x}+\frac{{{x}}^{{2}}}{{{2}!}}+\frac{{{x}}^{{3}}}{{{3}!}}+\frac{{{x}}^{{4}}}{{{4}!}}+\frac{{{x}}^{{5}}}{{{5}!}}+$ ...

To apply this on the given integral $\displaystyle{\int_{{0}}^{{1}}}{{e}}^{{-{{x}}^{{2}}}}{\left.{d}{x}\right.}$ ,

we replace the " $\displaystyle{x}$ " with " $\displaystyle-{{x}}^{{2}}$ ".

$\displaystyle{{e}}^{{-{{x}}^{{2}}}}={\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{{x}}^{{2}}\right)}}^{{n}}}{{{n}!}}$

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

$\displaystyle={1}-{{x}}^{{2}}+\frac{{{x}}^{{4}}}{{2}}-\frac{{{x}}^{{6}}}{{6}}+\frac{{{x}}^{{8}}}{{24}}-\frac{{{x}}^{{{10}}}}{{120}}+\frac{{{x}}^{{{12}}}}{{{6}!}}-$ ...

The integral becomes:

$\displaystyle{\int_{{0}}^{{1}}}{{e}}^{{-{{x}}^{{2}}}}{\left.{d}{x}\right.}={\int_{{0}}^{{1}}}{\left[{1}-{{x}}^{{2}}+\frac{{{x}}^{{4}}}{{2}}-\frac{{{x}}^{{6}}}{{6}}+\frac{{{x}}^{{8}}}{{24}}-\frac{{{x}}^{{{10}}}}{{120}}+\frac{{{x}}^{{{12}}}}{{720}}-\ldots\right]}{\left.{d}{x}\right.}$

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

$\displaystyle{\int_{{0}}^{{1}}}{\left[{1}-{{x}}^{{2}}+\frac{{{x}}^{{4}}}{{2}}-\frac{{{x}}^{{6}}}{{6}}+\frac{{{x}}^{{8}}}{{24}}-\frac{{{x}}^{{{10}}}}{{120}}+\frac{{{x}}^{{{12}}}}{{720}}-\ldots\right]}{\left.{d}{x}\right.}$

$\displaystyle={\left[{x}-\frac{{{x}}^{{3}}}{{3}}+\frac{{{x}}^{{5}}}{{{2}\cdot{5}}}-\frac{{{x}}^{{7}}}{{{6}\cdot{7}}}+\frac{{{x}}^{{9}}}{{{24}\cdot{9}}}-\frac{{{x}}^{{{11}}}}{{{120}\cdot{11}}}+\frac{{{x}}^{{{13}}}}{{{720}\cdot{13}}}-\ldots\right]}{{\mid}_{{0}}^{{1}}}$

$\displaystyle={\left[{x}-\frac{{{x}}^{{3}}}{{3}}+\frac{{{x}}^{{5}}}{{10}}-\frac{{{x}}^{{7}}}{{42}}+\frac{{{x}}^{{9}}}{{216}}-\frac{{{x}}^{{{11}}}}{{1320}}+\frac{{{x}}^{{{13}}}}{{9360}}-\ldots\right]}{{\mid}_{{0}}^{{1}}}$

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

$\displaystyle{F}{\left({1}\right)}={1}-\frac{{{1}}^{{3}}}{{3}}+\frac{{{1}}^{{5}}}{{10}}-\frac{{{1}}^{{7}}}{{42}}+\frac{{{1}}^{{9}}}{{216}}-\frac{{{1}}^{{{11}}}}{{1320}}+\frac{{{1}}^{{{13}}}}{{9360}}-$ ...

$\displaystyle={1}-\frac{{1}}{{3}}+\frac{{1}}{{10}}-\frac{{1}}{{42}}+\frac{{1}}{{216}}-\frac{{1}}{{1320}}+\frac{{1}}{{9360}}-$ ...

$\displaystyle{F}{\left({0}\right)}={0}-\frac{{{0}}^{{3}}}{{3}}+\frac{{{0}}^{{5}}}{{10}}-\frac{{{0}}^{{7}}}{{42}}+\frac{{{0}}^{{9}}}{{216}}-\frac{{{0}}^{{{11}}}}{{1320}}+\frac{{{0}}^{{{13}}}}{{9360}}-$ ...

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

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

We can stop at 7th term $\displaystyle{\left(\frac{{1}}{{9360}}\approx{0.0001068}\right)}$ since we only need error less than 0.0001.

Then,

$\displaystyle{F}{\left({1}\right)}-{F}{\left({0}\right)}={\left[{1}-\frac{{1}}{{3}}+\frac{{1}}{{10}}-\frac{{1}}{{42}}+\frac{{1}}{{216}}-\frac{{1}}{{1320}}+\frac{{1}}{{9360}}\right]}-{\left[{0}\right]}$

$\displaystyle={0.7468360343}$

Thus, the approximation of the integral will be:

$\displaystyle{\int_{{0}}^{{1}}}{{e}}^{{-{{x}}^{{2}}}}{\left.{d}{x}\right.}\approx{0.7468}$

New Notes Blog

Saturday, 26 September 2015

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

Saturday, 26 September 2015

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

No comments:

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?

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