`int_3^6 1 / (25 + (x-3)^2) dx` Evaluate the definite integral

Wednesday, August 3, 2011

$\displaystyle{\int_{{3}}^{{6}}}\frac{{1}}{{{25}+{{\left({x}-{3}\right)}}^{{2}}}}{\left.{d}{x}\right.}$ Evaluate the definite integral

To be able to solve for definite integral, we follow the first fundamental theorem of calculus: $\displaystyle{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

such that f is continuous and F is the antiderivative of f in a closed interval $\displaystyle{\left[{a},{b}\right]}$ .

The $\displaystyle{\left[{a}.{b}\right]}$ is the boundary limits of the integral such as lower bound=a and upper bound = b.

For the given problem: $\displaystyle{\int_{{3}}^{{{6}}}}\frac{{1}}{{{25}+{{\left({x}-{3}\right)}}^{{2}}}}{\left.{d}{x}\right.}$ ,

it resembles the basic integration formula:

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

By comparison: $\displaystyle\frac{{{d}{u}}}{{{{a}}^{{2}}+{{u}}^{{2}}}}{v}{s}{\left(\frac{{1}}{{{25}+{{\left({x}-{3}\right)}}^{{2}}}}\right)}{\left.{d}{x}\right.}$ , we may apply

u-substitution by letting:

$\displaystyle{{u}}^{{2}}={{\left({x}-{3}\right)}}^{{2}}$ then $\displaystyle{u}={x}-{3}$

where $\displaystyle{{a}}^{{2}}={25}{\quad\text{or}\quad}{{5}}^{{2}}$ then $\displaystyle{a}={5}$

Derivative of u will be $\displaystyle{d}{u}={1}{\left.{d}{x}\right.}$ or $\displaystyle{d}{u}={\left.{d}{x}\right.}$ .

$\displaystyle{\int_{{3}}^{{{6}}}}\frac{{1}}{{{25}+{{\left({x}-{3}\right)}}^{{2}}}}{\left.{d}{x}\right.}={\int_{{3}}^{{{6}}}}\frac{{1}}{{{25}+{{\left({u}\right)}}^{{2}}}}{d}{u}$

Applying the formula:

$\displaystyle{\int_{{3}}^{{{6}}}}\frac{{1}}{{{25}+{{\left({u}\right)}}^{{2}}}}{d}{u}={\left(\frac{{1}}{{5}}\right)}{\arctan{{\left(\frac{{u}}{{5}}\right)}}}{{\mid}_{{3}}^{{6}}}$

Plug-in $\displaystyle{u}={x}-{3}$ to express the indefinite integral in terms of x:

$\displaystyle{\left(\frac{{1}}{{5}}\right)}{\arctan{{\left(\frac{{u}}{{5}}\right)}}}{{\left|_{{3}}^{{6}}={\left(\frac{{1}}{{5}}\right)}{\arctan{{\left(\frac{{{x}-{3}}}{{5}}\right)}}}\right|}_{{3}}^{{6}}}$

Recall $\displaystyle{F}{\left({x}\right)}{{\mid}_{{a}}^{{b}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$ then:

$\displaystyle{\left(\frac{{1}}{{5}}\right)}{\arctan{{\left(\frac{{{x}-{3}}}{{5}}\right)}}}{{\mid}_{{3}}^{{6}}}={F}{\left({6}\right)}-{F}{\left({3}\right)}$

$\displaystyle={\left(\frac{{1}}{{5}}\right)}{\arctan{{\left(\frac{{{6}-{3}}}{{5}}\right)}}}-{\left(\frac{{1}}{{5}}\right)}{\arctan{{\left(\frac{{{3}-{3}}}{{5}}\right)}}}$

$\displaystyle={\left(\frac{{1}}{{5}}\right)}{\arctan{{\left(\frac{{3}}{{5}}\right)}}}-{\left(\frac{{1}}{{5}}\right)}{\arctan{{\left(\frac{{0}}{{5}}\right)}}}$

$\displaystyle={\left(\frac{{1}}{{5}}\right)}{\arctan{{\left(\frac{{3}}{{5}}\right)}}}-{0}$

$\displaystyle={\left(\frac{{1}}{{5}}\right)}{\arctan{{\left(\frac{{3}}{{5}}\right)}}}$ as the Final Answer.

Note: $\displaystyle{\arctan{{\left(\frac{{0}}{{5}}\right)}}}={\arctan{{\left({0}\right)}}}={0}$

since $\displaystyle{\tan{{\left(\theta\right)}}}={0}$ when $\displaystyle\theta={0}$

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Wednesday, August 3, 2011

$\displaystyle{\int_{{3}}^{{6}}}\frac{{1}}{{{25}+{{\left({x}-{3}\right)}}^{{2}}}}{\left.{d}{x}\right.}$ Evaluate the definite 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?

Wednesday, August 3, 2011

Evaluate the definite integral

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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?

$\displaystyle{\int_{{3}}^{{6}}}\frac{{1}}{{{25}+{{\left({x}-{3}\right)}}^{{2}}}}{\left.{d}{x}\right.}$ Evaluate the definite integral

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