Blog: `int_1^2 x^4 (ln(x))^2 dx` Evaluate the integral

Thursday, January 21, 2016

$\displaystyle{\int_{{1}}^{{2}}}{{x}}^{{4}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle{\int_{{1}}^{{2}}}{{x}}^{{4}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}{\left.{d}{x}\right.}$

If f(x) and g(x) are differentiable functions, then

$\displaystyle\int{f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}{\left.{d}{x}\right.}={f{{\left({x}\right)}}}{g{{\left({x}\right)}}}-\int{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}{\left.{d}{x}\right.}$

If we write f(x)=u and g'(x)=v, then

$\displaystyle\int{u}{v}{\left.{d}{x}\right.}={u}\int{v}{\left.{d}{x}\right.}-\int{\left({u}'\int{v}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}$

Using the above method of integration by parts,

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

$\displaystyle={{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}\cdot\frac{{{x}}^{{5}}}{{5}}-\int{\left({2}{\ln{{\left({x}\right)}}}\cdot\frac{{1}}{{x}}{\left(\frac{{{x}}^{{5}}}{{5}}\right)}\right)}{\left.{d}{x}\right.}$

$\displaystyle={{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}\frac{{{x}}^{{5}}}{{5}}-\frac{{2}}{{5}}\int{{x}}^{{4}}{\ln{{\left({x}\right)}}}{\left.{d}{x}\right.}$

again applying integration by parts,

$\displaystyle=\frac{{{x}}^{{5}}}{{5}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}-\frac{{2}}{{5}}{\left({\ln{{\left({x}\right)}}}\int{{x}}^{{4}}{\left.{d}{x}\right.}-\int{\left(\frac{{d}}{{\left.{d}{x}}}{\left({\ln{{\left({x}\right)}}}\right)}\int{{x}}^{{4}}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}\right)}$

$\displaystyle=\frac{{{x}}^{{5}}}{{5}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}-\frac{{2}}{{5}}{\left({\ln{{\left({x}\right)}}}\frac{{{x}}^{{5}}}{{5}}-\int{\left(\frac{{1}}{{x}}\cdot\frac{{{x}}^{{5}}}{{5}}\right)}{\left.{d}{x}\right.}\right)}$

$\displaystyle=\frac{{{x}}^{{5}}}{{5}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}-\frac{{2}}{{25}}{{x}}^{{5}}{\ln{{\left({x}\right)}}}+\frac{{2}}{{25}}\int{{x}}^{{4}}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{{x}}^{{5}}}{{5}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}-\frac{{2}}{{25}}{{x}}^{{5}}{\ln{{\left({x}\right)}}}+\frac{{2}}{{25}}\cdot\frac{{{x}}^{{5}}}{{5}}$

adding constant to the solution,

$\displaystyle=\frac{{{x}}^{{5}}}{{5}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}-\frac{{2}}{{25}}{{x}}^{{5}}{\ln{{\left({x}\right)}}}+\frac{{2}}{{125}}{{x}}^{{5}}+{C}$

Now evaluate the definite integral,

$\displaystyle{\int_{{1}}^{{2}}}{{x}}^{{4}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}{\left.{d}{x}\right.}={{\left[\frac{{{x}}^{{5}}}{{5}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}-\frac{{2}}{{25}}{{x}}^{{5}}{\ln{{\left({x}\right)}}}+\frac{{2}}{{125}}{{x}}^{{5}}\right]}_{{1}}^{{2}}}$

$\displaystyle={\left[\frac{{{2}}^{{5}}}{{5}}{{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}-\frac{{2}}{{25}}\cdot{{2}}^{{5}}{\ln{{\left({2}\right)}}}+\frac{{2}}{{125}}{\left({{2}}^{{5}}\right)}\right]}-{\left[\frac{{{1}}^{{5}}}{{5}}{{\left({\ln{{\left({1}\right)}}}\right)}}^{{2}}-\frac{{2}}{{25}}{{\left({1}\right)}}^{{5}}{\ln{{\left({1}\right)}}}+\frac{{2}}{{125}}{\left({{1}}^{{5}}\right)}\right]}$

$\displaystyle={\left[\frac{{32}}{{5}}{{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}-\frac{{64}}{{25}}{\ln{{\left({2}\right)}}}+\frac{{64}}{{125}}\right]}-{\left[\frac{{2}}{{125}}\right]}$

$\displaystyle=\frac{{32}}{{5}}{{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}-\frac{{64}}{{25}}{\ln{{\left({2}\right)}}}+\frac{{62}}{{125}}$

$\displaystyle{\int_{{1}}^{{2}}}{{x}}^{{4}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}{\left.{d}{x}\right.}$

If f(x) and g(x) are differentiable functions, then

If we write f(x)=u and g'(x)=v, then

$\displaystyle\int{u}{v}{\left.{d}{x}\right.}={u}\int{v}{\left.{d}{x}\right.}-\int{\left({u}'\int{v}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}$

Using the above method of integration by parts,

$\displaystyle={{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}\frac{{{x}}^{{5}}}{{5}}-\frac{{2}}{{5}}\int{{x}}^{{4}}{\ln{{\left({x}\right)}}}{\left.{d}{x}\right.}$

again applying integration by parts,

$\displaystyle=\frac{{{x}}^{{5}}}{{5}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}-\frac{{2}}{{25}}{{x}}^{{5}}{\ln{{\left({x}\right)}}}+\frac{{2}}{{25}}\int{{x}}^{{4}}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{{x}}^{{5}}}{{5}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}-\frac{{2}}{{25}}{{x}}^{{5}}{\ln{{\left({x}\right)}}}+\frac{{2}}{{25}}\cdot\frac{{{x}}^{{5}}}{{5}}$

adding constant to the solution,

$\displaystyle=\frac{{{x}}^{{5}}}{{5}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}-\frac{{2}}{{25}}{{x}}^{{5}}{\ln{{\left({x}\right)}}}+\frac{{2}}{{125}}{{x}}^{{5}}+{C}$

Now evaluate the definite integral,

$\displaystyle={\left[\frac{{32}}{{5}}{{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}-\frac{{64}}{{25}}{\ln{{\left({2}\right)}}}+\frac{{64}}{{125}}\right]}-{\left[\frac{{2}}{{125}}\right]}$

$\displaystyle=\frac{{32}}{{5}}{{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}-\frac{{64}}{{25}}{\ln{{\left({2}\right)}}}+\frac{{62}}{{125}}$

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Thursday, January 21, 2016

$\displaystyle{\int_{{1}}^{{2}}}{{x}}^{{4}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}{\left.{d}{x}\right.}$ Evaluate the integral

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What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Thursday, January 21, 2016

Evaluate the integral

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What is the Exposition, Rising Action, Climax, and Falling Action of &quot;One Thousand Dollars&quot;?

$\displaystyle{\int_{{1}}^{{2}}}{{x}}^{{4}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}{\left.{d}{x}\right.}$ Evaluate the integral

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?