`int_0^2 dx/(x^2-2x+2) ` Find or evaluate the integral by completing the square

Tuesday, January 28, 2014

$\displaystyle{\int_{{0}}^{{2}}}\frac{{\left.{d}{x}}}{{{{x}}^{{2}}-{2}{x}+{2}}}$ Find or evaluate the integral by completing the square

To be able to evaluate the given integral: $\displaystyle{\int_{{0}}^{{2}}}\frac{{{\left.{d}{x}\right.}}}{{{{x}}^{{2}}-{2}{x}+{2}}}$ , we

complete the square of the expression: $\displaystyle{{x}}^{{2}}-{2}{x}+{2}$ .

To complete the square, we add and subtract $\displaystyle{{\left(-\frac{{b}}{{{2}{a}}}\right)}}^{{2}}$ .

The $\displaystyle{{x}}^{{2}}-{2}{x}+{2}$ resembles the $\displaystyle{a}{{x}}^{{2}}+{b}{x}+{c}$ where:

$\displaystyle{a}={1}$ , $\displaystyle{b}=-{2}$ and $\displaystyle{c}={2}$ .

Then,

$\displaystyle{{\left(-\frac{{b}}{{{2}{a}}}\right)}}^{{2}}={{\left(-\frac{{-{2}}}{{{2}{\left({1}\right)}}}\right)}}^{{2}}$

$\displaystyle={{\left(\frac{{2}}{{2}}\right)}}^{{2}}$

$\displaystyle={{\left({1}\right)}}^{{2}}$

$\displaystyle={1}$

Add and subtract 1 :

$\displaystyle{{x}}^{{2}}-{2}{x}+{2}+{1}-{1}$

Rearrange as: $\displaystyle{\left({{x}}^{{2}}-{2}{x}+{1}\right)}+{2}-{1}={{\left({x}-{1}\right)}}^{{2}}+{1}$

Plug-in $\displaystyle{{x}}^{{2}}-{2}{x}+{2}={{\left({x}-{1}\right)}}^{{2}}+{1}$ in the given integral: $\displaystyle{\int_{{0}}^{{2}}}\frac{{{\left.{d}{x}\right.}}}{{{{x}}^{{2}}-{2}{x}+{2}}}$ .

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

This resembles the basic integral formula for inverse tangent function:

$\displaystyle\int\frac{{{d}{u}}}{{{{u}}^{{2}}+{{a}}^{{2}}}}=\frac{{1}}{{a}}\cdot{\arctan{{\left(\frac{{u}}{{a}}\right)}}}+{C}$

Then indefinite integral F(x)+C,

$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{{\left({x}-{1}\right)}}^{{2}}+{{1}}^{{2}}}}=\frac{{1}}{{1}}\cdot{\arctan{{\left(\frac{{{x}-{1}}}{{1}}\right)}}}+{C}$

$\displaystyle={\arctan{{\left({x}-{1}\right)}}}+{C}$

For the definite integral, we apply: $\displaystyle{F}{\left({x}\right)}{{\mid}_{{a}}^{{b}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$ .

$\displaystyle{\arctan{{\left({x}-{1}\right)}}}{{\mid}_{{0}}^{{2}}}={\arctan{{\left({2}-{1}\right)}}}-{\arctan{{\left({0}-{1}\right)}}}$

$\displaystyle={\arctan{{\left({1}\right)}}}-{\arctan{{\left(-{1}\right)}}}$

$\displaystyle=\frac{\pi}{{4}}-{\left(-\frac{\pi}{{4}}\right)}$

$\displaystyle=\frac{\pi}{{4}}+\frac{\pi}{{4}}$

$\displaystyle=\frac{{{2}\pi}}{{4}}$

$\displaystyle=\frac{\pi}{{2}}$

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Tuesday, January 28, 2014

$\displaystyle{\int_{{0}}^{{2}}}\frac{{\left.{d}{x}}}{{{{x}}^{{2}}-{2}{x}+{2}}}$ Find or evaluate the integral by completing the square

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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, January 28, 2014

Find or evaluate the integral by completing the square

No comments:

Post a Comment

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

$\displaystyle{\int_{{0}}^{{2}}}\frac{{\left.{d}{x}}}{{{{x}}^{{2}}-{2}{x}+{2}}}$ Find or evaluate the integral by completing the square

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