<s>[INST] 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.
 [/INST] The given condition $y\le x$and $x^2\le y\le1$ implies $x^2\le y\le x$. 
Since $x^2\le x$, $0\le x\le 1$.
$$\int_{0}^1\int_{x^2}^x\frac{21}{4}x^2ydydx=\int_{0}^1\left[\frac{21}{8}x^2y^2\right]^x_{x^2}dx=\int_{0}^1\frac{21}{8}x^2(x^2-x^4)=\frac{21}{8}\left[\frac{x^5}{5}-\frac{x^7}{7}\right]^1_{0}=\frac{3}{20}$$
</s>