`int 2 / sqrt(-x^2+4x) dx` Find or evaluate the integral by completing the square

Monday, September 27, 2010

$\displaystyle\int\frac{{2}}{\sqrt{{-{{x}}^{{2}}+{4}{x}}}}{\left.{d}{x}\right.}$ Find or evaluate the integral by completing the square

To evaluate the given integral: $\displaystyle\int\frac{{2}}{\sqrt{{-{{x}}^{{2}}+{4}{x}}}}{\left.{d}{x}\right.}$ , we may apply the basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

The integral becomes:

$\displaystyle{2}\int\frac{{\left.{d}{x}}}{\sqrt{{-{{x}}^{{2}}+{4}{x}}}}$

We complete the square for the expression $\displaystyle{\left(-{{x}}^{{2}}+{4}{x}\right)}$ .

Completing the square:

For the first step, factor out (-1): $\displaystyle{\left(-{{x}}^{{2}}+{4}{x}\right)}={\left(-{1}\right)}{\left({{x}}^{{2}}-{4}{x}\right)}{\quad\text{or}\quad}-{\left({{x}}^{{2}}-{4}{x}\right)}$

The $\displaystyle{{x}}^{{2}}-{4}{x}$ or $\displaystyle{{x}}^{{-{{4}}}}{x}+{0}$ resembles the $\displaystyle{a}{{x}}^{{2}}+{b}{x}+{c}$ where:

$\displaystyle{a}={1}$ , $\displaystyle{b}=-{4}$ and $\displaystyle{c}={0}$ .

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

Using $\displaystyle{a}={1}$ and $\displaystyle{b}=-{4}$ , we get:

$\displaystyle{{\left(-\frac{{b}}{{{2}{a}}}\right)}}^{{2}}={{\left(-\frac{{-{4}}}{{{2}{\left({1}\right)}}}\right)}}^{{2}}$

$\displaystyle={{\left(\frac{{4}}{{2}}\right)}}^{{2}}$

$\displaystyle={{2}}^{{2}}$

$\displaystyle={4}$

Add and subtract 4 inside the $\displaystyle{\left({{x}}^{{2}}-{4}{x}\right)}$ :

$\displaystyle-{\left({{x}}^{{2}}-{4}{x}+{4}-{4}\right)}$

Distribute the negative sign on -4 to rewrite it as:

$\displaystyle-{\left({{x}}^{{2}}-{4}{x}+{4}\right)}+{4}$

Factor the perfect square trinomial: $\displaystyle{{x}}^{{2}}-{4}{x}+{4}={{\left({x}-{2}\right)}}^{{2}}$ .

$\displaystyle-{{\left({x}-{2}\right)}}^{{2}}+{4}$

For the original problem, we let: $\displaystyle-{{x}}^{{2}}+{4}{x}=-{{\left({x}-{2}\right)}}^{{2}}+{4}$ :

$\displaystyle{2}\int\frac{{\left.{d}{x}}}{\sqrt{{-{{x}}^{{2}}+{4}{x}}}}={2}\int\frac{{\left.{d}{x}}}{\sqrt{{-{{\left({x}-{2}\right)}}^{{2}}+{4}}}}$

It can also be rewritten as:

$\displaystyle{2}\int\frac{{\left.{d}{x}}}{\sqrt{{-{{\left({x}-{2}\right)}}^{{2}}+{{2}}^{{2}}}}}={2}\int\frac{{\left.{d}{x}}}{\sqrt{{{{2}}^{{2}}-{{\left({x}-{2}\right)}}^{{2}}}}}$

The integral part resembles the integral formula:

$\displaystyle\int\frac{{{d}{u}}}{\sqrt{{{{a}}^{{2}}-{{u}}^{{2}}}}}={\arcsin{{\left(\frac{{u}}{{a}}\right)}}}+{C}$ .

Applying the formula, we get:

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

Then the indefinite integral :

$\displaystyle\int\frac{{2}}{\sqrt{{-{{x}}^{{2}}+{4}{x}}}}{\left.{d}{x}\right.}={2}{\arcsin{{\left(\frac{{{x}-{2}}}{{2}}\right)}}}+{C}$

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Monday, September 27, 2010

$\displaystyle\int\frac{{2}}{\sqrt{{-{{x}}^{{2}}+{4}{x}}}}{\left.{d}{x}\right.}$ 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?

Monday, September 27, 2010

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?

$\displaystyle\int\frac{{2}}{\sqrt{{-{{x}}^{{2}}+{4}{x}}}}{\left.{d}{x}\right.}$ 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?