Answer
a. $\int\int_{R^2}f(x,y)dA=1$
b. $P(X\geq\frac{1}2)=\frac{3}4
\\P(X\geq\frac{1}2,Y\leq\frac{1}2)=\frac{3}{16}$
c. $E(X)=E(Y)=\frac{2}3$
Work Step by Step
a,$$\int\int_{R^2}f(x,y)dA=\int_0^1\int_0^14xydxdy
\\=\int_0^12xdx\int_0^12ydy
\\=\biggr[x^2\biggr]_0^1\biggr[y^2\biggr]_0^1
\\=(1)(1)
\\=1$$
and $4xy\geq0$ for $0\leq x\leq1, 0\leq y\leq1$ therefore $f(x,y)$ is a joint density function.
b,(i)$$P(X\geq \frac{1}2)=\int_0^1\int_{\frac{1}2}^14xydxdy
\\=\int_0^12ydy\int_{\frac{1}2}^12xdx
\\=\biggr[y^2\biggr]_0^1\biggr[x^2\biggr]_{\frac{1}2}^1
\\=(1)\biggr[1^2-\biggr(\frac{1}2\biggr)^2\biggr]
\\=\frac{3}4$$
(ii),$$P(X\geq\frac{1}2,Y\leq\frac{1}2)=\int_0^{\frac{1}2}\int_{\frac{1}2}^14xydxdy
\\=\int_0^{\frac{1}2}2ydy\int_{\frac{1}2}^12xdx
\\=\biggr[y^2\biggr]_0^{\frac{1}2}\biggr[x^2\biggr]_{\frac{1}2}^1
\\=\biggr(\frac{1}2\biggr)^2\biggr[1^2-\biggr(\frac{1}2\biggr)^2\biggr]
\\=\frac{3}{16}$$
c,$$E(X)=\int\int_{R^2}xf(x,y)dA=\int_0^1\int_{0}^1 4x^2ydxdy
\\=\int_0^12ydy\int_{0}^1 2x^2dx
\\=\biggr[y^2\biggr]_0^1\biggr[\frac{2}3 x^3\biggr]_0^1
\\=\frac{2}{3}$$
$$E(Y)=\int\int_{R^2}yf(x,y)dA=\int_0^1\int_{0}^1 4xy^2dxdy
\\=\int_0^12y^2dy\int_{0}^1 2xdx
\\=\biggr[\frac{2}3 y^3\biggr]_0^1\biggr[x^2\biggr]_0^1
\\=\frac{2}{3}$$
