Notes: `int_0^3 sqrt(x^2+16) dx` Use integration tables to evaluate the definite integral.

Saturday, 18 October 2014

$\displaystyle{\int_{{0}}^{{3}}}\sqrt{{{{x}}^{{2}}+{16}}}{\left.{d}{x}\right.}$ Use integration tables to evaluate the definite integral.

Recall indefinite integral follows $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

where:

$\displaystyle{f{{\left({x}\right)}}}$ as the integrand

$\displaystyle{F}{\left({x}\right)}$ as the antiderivative of f(x)

$\displaystyle{C}$ as the constant of integration.

From the table of integrals, we follow the formula:

$\displaystyle\sqrt{{{{x}}^{{2}}\pm{{a}}^{{2}}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{x}\sqrt{{{{x}}^{{2}}\pm{{a}}^{{2}}}}\pm\frac{{1}}{{2}}{{a}}^{{2}}{\ln}{\left|{x}+\sqrt{{{{x}}^{{2}}\pm{{a}}^{{2}}}}\right|}$

From the given problem $\displaystyle{\int_{{0}}^{{3}}}\sqrt{{{{x}}^{{2}}+{16}}}{\left.{d}{x}\right.}$ , we have a addition sign (+) in between terms inside the square root sign. Then, we follow the formula:

$\displaystyle\int\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{x}\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}+\frac{{1}}{{2}}{{a}}^{{2}}{\ln}{\left|{x}+\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}\right|}$

Take note that we...

Recall indefinite integral follows $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

where:

$\displaystyle{f{{\left({x}\right)}}}$ as the integrand

$\displaystyle{F}{\left({x}\right)}$ as the antiderivative of f(x)

$\displaystyle{C}$ as the constant of integration.

From the table of integrals, we follow the formula:

Take note that we can express $\displaystyle{16}={{4}}^{{2}}$ then the given problem becomes: $\displaystyle{\int_{{0}}^{{3}}}\sqrt{{{{x}}^{{2}}+{{4}}^{{2}}}}{\left.{d}{x}\right.}$ .

The $\displaystyle{{x}}^{{2}}+{{4}}^{{2}}$ resembles the $\displaystyle{{x}}^{{2}}+{{a}}^{{2}}$ in the formula. Then by comparison, the corresponding values are: $\displaystyle{x}={x}$ and $\displaystyle{a}={4}.$

Plug-in $\displaystyle{x}={x}$ and $\displaystyle{a}={4}$ on the formula, we get:

$\displaystyle{\int_{{0}}^{{3}}}\sqrt{{{{x}}^{{2}}+{16}}}{\left.{d}{x}\right.}$ $\displaystyle={\left[\frac{{1}}{{2}}{x}\sqrt{{{{x}}^{{2}}+{{4}}^{{2}}}}+\frac{{1}}{{2}}\cdot{{4}}^{{2}}{\ln}{\left|{x}+\sqrt{{{{x}}^{{2}}+{{4}}^{{2}}}}\right|}\right]}{{\mid}_{{0}}^{{3}}}$

$\displaystyle={\left[\frac{{1}}{{2}}{x}\sqrt{{{{x}}^{{2}}+{16}}}+\frac{{1}}{{2}}\cdot{16}{\ln}{\left|{x}+\sqrt{{{{x}}^{{2}}+{16}}}\right|}\right]}{{\mid}_{{0}}^{{3}}}$

$\displaystyle={\left[\frac{{1}}{{2}}{x}\sqrt{{{{x}}^{{2}}+{16}}}+{8}{\ln}{\left|{x}+\sqrt{{{{x}}^{{2}}+{16}}}\right|}\right]}{{\mid}_{{0}}^{{3}}}$

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

$\displaystyle{\left[\frac{{1}}{{2}}{x}\sqrt{{{{x}}^{{2}}+{16}}}+{8}{\ln}{\left|{x}+\sqrt{{{{x}}^{{2}}+{16}}}\right|}\right]}{{\mid}_{{0}}^{{3}}}$

$\displaystyle={\left[\frac{{1}}{{2}}\cdot{3}\sqrt{{{{3}}^{{2}}+{16}}}+{8}{\ln}{\left|{3}+\sqrt{{{{3}}^{{2}}+{16}}}\right|}\right]}-{\left[\frac{{1}}{{2}}\cdot{0}\sqrt{{{{0}}^{{2}}+{16}}}+{8}{\ln}{\left|{0}+\sqrt{{{{0}}^{{2}}+{16}}}\right|}\right]}$

$\displaystyle={\left[\frac{{3}}{{2}}\sqrt{{{9}+{16}}}+{8}{\ln}{\left|{3}+\sqrt{{{9}+{16}}}\right|}\right]}-{\left[{0}\cdot\sqrt{{{0}+{16}}}+{8}{\ln}{\left|{0}+\sqrt{{{0}+{16}}}\right|}\right]}$

$\displaystyle={\left[\frac{{3}}{{2}}\cdot{5}+{8}{\ln}{\left|{3}+{5}\right|}\right]}-{\left[{0}\cdot{4}+{8}{\ln}{\left|{0}+{4}\right|}\right]}$

$\displaystyle={\left[\frac{{15}}{{2}}+{8}{\ln}{\left|{8}\right|}\right]}-{\left[{0}+{8}{\ln}{\left|{4}\right|}\right]}$

$\displaystyle=\frac{{15}}{{2}}+{8}{\ln}{\left|{8}\right|}-{0}-{8}{\ln}{\left|{4}\right|}$

$\displaystyle=\frac{{15}}{{2}}+{8}{\ln}{\left|{8}\right|}-{8}{\ln}{\left|{4}\right|}$

$\displaystyle=\frac{{15}}{{2}}+{8}{\left({\ln}{\left|{8}\right|}-{\ln}{\left|{4}\right|}\right)}$

Apply natural logarithm property: $\displaystyle{\ln{{\left({x}\right)}}}-{\ln{{\left({y}\right)}}}={\ln{{\left(\frac{{x}}{{y}}\right)}}}$ .

$\displaystyle=\frac{{15}}{{2}}+{8}{\ln}{\left|\frac{{8}}{{4}}\right|}$

$\displaystyle=\frac{{15}}{{2}}+{8}{\ln}{\left|{2}\right|}$

Apply natural logarithm property: $\displaystyle{n}\cdot{\ln{{\left({x}\right)}}}={\ln{{\left({{x}}^{{n}}\right)}}}$ .

$\displaystyle=\frac{{15}}{{2}}+{\ln}{\left|{{2}}^{{8}}\right|}$

= $\displaystyle\frac{{15}}{{2}}+{\ln}{\left|{256}\right|}$ or $\displaystyle{13.05}$ ( approximated value)

Notes

Saturday, 18 October 2014

$\displaystyle{\int_{{0}}^{{3}}}\sqrt{{{{x}}^{{2}}+{16}}}{\left.{d}{x}\right.}$ Use integration tables to evaluate the definite integral.

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Saturday, 18 October 2014

Use integration tables to evaluate the definite integral.

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Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle{\int_{{0}}^{{3}}}\sqrt{{{{x}}^{{2}}+{16}}}{\left.{d}{x}\right.}$ Use integration tables to evaluate the definite integral.

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?