# How to have this definite integral?

Suppose that a function $f$ of $x$ and $y$ be defined as follows:$$f(x,y) = \begin{cases} \frac{21}{4}x^2y & \text{for x^2 \leq y\leq 1,} \\ 0 & \text{otherwise. } \\ \end{cases}$$ I have to determine the value of integral for which $y\leq x$ also holds.

The answer is $\frac{3}{20}$ and I also get it using figure for $x$ and $y$, but don't know how to get it with calculations.

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The given condition $y\le x$and $x^2\le y\le1$ implies $x^2\le y\le x$.

Since $x^2\le x$, $0\le x\le 1$.

$$\int_{0}^1\int_{x^2}^x\frac{21}{4}x^2ydydx=\int_{0}^1\left[\frac{21}{8}x^2y^2\right]^x_{x^2}dx=\int_{0}^1\frac{21}{8}x^2(x^2-x^4)=\frac{21}{8}\left[\frac{x^5}{5}-\frac{x^7}{7}\right]^1_{0}=\frac{3}{20}$$

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Nice, but why $x^2<y<x$ and $x^2<y<x$ only? – Silent Oct 13 '13 at 1:15
Oh, many many thanks. – Silent Oct 13 '13 at 1:24

We find $$\int_{x=0}^1 \left(\int_{y=x^2}^{x}\frac{21}{4}x^2y\,dy\right)\,dx.$$ The integration with respect to $y$ gives $\frac{21}{8}x^2(x^2-x^4)$. Integration with respect to $x$ yields $\frac{21}{8} \left(\frac{1}{5}-\frac{1}{7}\right)$. Simplify.

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Sir, how did you choose values $x^2<y<x$ and $0<x<1$ ? – Silent Oct 13 '13 at 1:13
You had explicitly specified that the function was $x^2y$ for $0\le x^2\le y$, and $0$ elsewhere. You wanted the integral over the region where $y\le x$. That means we are integrating $x^2y$ with respect to $y$, $y=x^2$ to $y=x$. The part $x^2\le y \le x$ forces $x$ to travel from $0$ to $1$. A picture of the region of integration would help, one should automatically draw one in this kind of problem. . – André Nicolas Oct 13 '13 at 1:33
Thank you ,sir. – Silent Oct 13 '13 at 1:41


\begin{align} {\cal I} & \equiv \int_{-\infty}^{\infty}{\rm d}x\int_{-\infty}^{\infty}{\rm d}y\, {21 \over 4}\,x^{2}y\, \Theta\pars{y - x^{2}}\Theta\pars{1 - y}\Theta\pars{x - y} \\[3mm]&= {21 \over 4}\int_{-\infty}^{1}{\rm d}y\,y\quad \overbrace{\int_{-\infty}^{\infty}{\rm d}x\, x^{2}\,\Theta\pars{y - x^{2}}\Theta\pars{x - y}}^{\equiv\ {\cal J}} \end{align}
\begin{align} {\cal J} &= -\,{1 \over 3}\int_{-\infty}^{\infty}{\rm d}x\, x^{3}\,\bracks{% -2x\,\delta\pars{y - x^{2}}\Theta\pars{x - y} + \Theta\pars{y - x^{2}}\delta\pars{x - y}} \\[3mm]&= {2 \over 3}\int_{-\infty}^{\infty}{\rm d}x\,x^{4}\,\Theta\pars{y}\, {\delta\pars{x - y^{1/2}} \over 2\verts{x}}\,\Theta\pars{y^{1/2} - y} - {1 \over 3}\,y^{3}\,\Theta\pars{y - y^{2}} \\[3mm]&= {1 \over 3}y^{3/2}\,\Theta\pars{y}\,\Theta\pars{y\bracks{1 -y}} - {1 \over 3}\,y^{3}\,\Theta\pars{y - y^{2}} = {1 \over 3}\pars{y^{3/2} - y^{3}}\Theta\pars{y}\Theta\pars{1 - y} \end{align}
\begin{align} {\cal I} &= {7 \over 4}\int_{0}^{1}\pars{y^{5/2} - y^{4}}\,{\rm d}y = {7 \over 4}\pars{{2 \over 7} - {1 \over 5}} = \color{#ff0000}{\large{3 \over 20}} \end{align}
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