Blog: `int_4^9 ln(y)/sqrt(y) dy` Evaluate the integral

Wednesday, June 1, 2016

$\displaystyle{\int_{{4}}^{{9}}}\frac{{\ln{{y}}}}{\sqrt{{y}}}{\left.{d}{y}\right.}$

To evaluate, apply integration by parts $\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$ .

So let:

$\displaystyle{u}={\ln{{y}}}$

and

$\displaystyle{d}{v}=\int\frac{{1}}{\sqrt{{y}}}{\left.{d}{y}\right.}$

Then, differentiate u and integrate dv.

$\displaystyle{d}{u}=\frac{{1}}{{y}}{\left.{d}{y}\right.}$

and

$\displaystyle{v}=\int\frac{{1}}{\sqrt{{y}}}{\left.{d}{y}\right.}=\int{{y}}^{{-\frac{{1}}{{2}}}}={2}{{y}}^{{\frac{{1}}{{2}}}}$

Plug-in them to the formula. So the integral becomes:

$\displaystyle\int\frac{{\ln{{y}}}}{{\sqrt{{y}}}}{\left.{d}{y}\right.}$

$\displaystyle={\ln{{y}}}\cdot{2}{{y}}^{{\frac{{1}}{{2}}}}-{2}{{y}}^{{\frac{{1}}{{2}}}}\cdot\frac{{1}}{{y}}{\left.{d}{y}\right.}$

$\displaystyle={2}{{y}}^{{\frac{{1}}{{2}}}}{\ln{{y}}}-{2}\int{{y}}^{{-\frac{{1}}{{2}}}}{\left.{d}{y}\right.}$

$\displaystyle={2}{{y}}^{{\frac{{1}}{{2}}}}{\ln{{y}}}-{2}\cdot{2}{{y}}^{{\frac{{1}}{{2}}}}$

$\displaystyle={2}\sqrt{{y}}{\ln{{y}}}-{4}\sqrt{{y}}$

And, substitute the limits of the integral.

$\displaystyle{\int_{{4}}^{{9}}}\frac{{\ln{{y}}}}{{y}}{\left.{d}{y}\right.}$

$\displaystyle={\left({2}\sqrt{{y}}{\ln{{y}}}-{4}\sqrt{{y}}\right)}{{\mid}_{{4}}^{{9}}}$

$\displaystyle={\left({2}\sqrt{{9}}{\ln{{9}}}-{4}\sqrt{{9}}\right)}-{\left({2}\sqrt{{4}}{\ln{{4}}}-{4}\sqrt{{4}}\right)}$

$\displaystyle={\left({2}\cdot{3}{\ln{{9}}}-{4}\cdot{3}\right)}-{\left({2}\cdot{2}{\ln{{4}}}-{4}\cdot{2}\right)}$

$\displaystyle={\left({6}{\ln{{9}}}-{12}\right)}-{\left({4}{\ln{{4}}}-{8}\right)}$

$\displaystyle={6}{\ln{{9}}}-{4}{\ln{{4}}}-{4}$

$\displaystyle={\ln{{\left({{9}}^{{6}}\right)}}}-{\ln{{\left({{4}}^{{4}}\right)}}}-{4}$

$\displaystyle={\ln{{\left(\frac{{{9}}^{{6}}}{{{4}}^{{4}}}\right)}}}-{4}$

$\displaystyle={\ln{{\left(\frac{{{{\left({{3}}^{{2}}\right)}}^{{6}}}}{{{{4}}^{{4}}}}\right)}}}-{4}$

$\displaystyle={\ln{{\left(\frac{{{{\left({{3}}^{{3}}\right)}}^{{4}}}}{{{{4}}^{{4}}}}\right)}}}-{4}$

$\displaystyle={\ln{{\left({{\left(\frac{{{3}}^{{3}}}{{4}}\right)}}^{{4}}\right)}}}-{4}$

$\displaystyle={4}{\ln{{\left(\frac{{27}}{{4}}\right)}}}-{4}$

Therefore, $\displaystyle{\int_{{4}}^{{9}}}\frac{{\ln{{y}}}}{\sqrt{{y}}}{\left.{d}{y}\right.}={4}{\ln{{\left(\frac{{27}}{{4}}\right)}}}-{4}$ .

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