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I need to integrate the following equation without using trigonometric substitution (which we haven't learned yet but I have been told would be the normal way to integrate it).

$\int_0^2 (y+8)\sqrt {-y^2+4y}$ $dx$

I know that the answer to this is $ 10\pi−\frac{8}{3} $ but I don't know how to demonstrate this.

I know that it's useful that the integral of $\sqrt {-y^2+4y}$ is just [a fourth of] the area of a circle with radius 2 but unless I can isolate the radical from (y+8) I can't use this fact. Thoughts?

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  • $\begingroup$ @FelixMarin Yes sorry, it was a formatting error. The bar was meant to extend over the whole equation. $\endgroup$ – Wynne Plaga Feb 2 '17 at 1:22
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Hint:

$$(y+8)\sqrt{-y^2+4y}$$

$$=(-\frac{1}{2}(-2y)+8)\sqrt{-y^2+4y}$$

$$=(-\frac{1}{2}(-2y+4)+10) \sqrt{-y^2+4y}$$

$$=-\frac{1}{2}(-2y+4)\sqrt{-y^2+4y}+10 \sqrt{-y^2+4y}$$

$\int_{0}^{2} \sqrt{-y^2+4y} \, dy$ represents a fourth of the area of a circle of radius $2$.

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  • $\begingroup$ Wonderfully done! +1 $\endgroup$ – Simply Beautiful Art Feb 2 '17 at 1:20
  • $\begingroup$ Thank you ! @SimplyBeautifulArt $\endgroup$ – Ahmed S. Attaalla Feb 2 '17 at 1:20
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Use Euler Substitution $\ds{y = {\ic t^{2} \over 2\pars{t + 2\ic}}}$ with $\ds{t = \root{-y^{2} + 4y} - y\ic}$. You'll get

\begin{align} &\int_{0}^{2}\pars{y + 8}\root{-y^{2} + 4y}\,\dd y \\[5mm] = &\ \int_{0}^{2 - 2\ic}\bracks{-5 + 2\ic t - {t^{2} \over 8} + {8 \over \pars{t + 2\ic}^{4}} + {40\ic \over \pars{t + 2\ic}^{3}} + {2 \over \pars{t + 2\ic}^{2}} + {20\ic \over t + 2\ic}}\,\dd t \end{align} which involves straightforward integrations. The answer is $\bbx{\ds{10\pi - {8 \over 3}}}$.

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Express $y+8$as a differential of $-y^2+4y$ and some constant.Then take $-y^2+4y$ as $t$ and proceed.what i meant was write $y+8=a(\frac{d}{dx} [-y^2+4]+b$.Then solve for a and b and put in the integral.

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  • $\begingroup$ Thanks for the help, however I'm afraid I'm not able to grasp your answer properly. What do you mean by >Express y+8y+8as a differential of −y2+4y−y2+4y and some constant. In case it's helpful, the two primary process of integration that I'm familiar with are u-sub and integration by parts, but my knowledge beyond these techniques is limited. $\endgroup$ – Wynne Plaga Feb 2 '17 at 1:10
  • $\begingroup$ @der_Fidelis check the edit $\endgroup$ – Navin Feb 2 '17 at 2:04

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