\(\displaystyle \int_{0}^{1}\int_{x}^{x^2} xy^2 dydx=\int_{0}^{1}\Big(\int_{x}^{x^2} xy^2 dy\Big)dx\)

Lets deal with the inner integral first.

\(\displaystyle \int_{x}^{x^2}xy^2 dy\)

\(\displaystyle \int_{x}^{x^2}xy^2 dy=\frac{1}{3}y^3x\Big|_{x}^{x^2}\)

\(\displaystyle =(\frac{1}{3}(x^2)^3\cdot x)-(\frac{1}{3}x^3\cdot x)\)

\(\displaystyle =\frac{1}{3}x^7-\frac{1}{3}x^4\)

Now we evaluate this expression in the outer integral.

\(\displaystyle \int_{0}^{1} \frac{1}{3}x^7-\frac{1}{3}x^4dx\)

\(\displaystyle =\frac{1}{3}\int_{0}^{1} x^7-x^4dx\)

\(\displaystyle =\frac{1}{3}\Big( \frac{1}{8}x^8-\frac{1}{5}x^5\Big)\Big|_{0}^{1}\)

\(\displaystyle =\frac{1}{3}\Big(\frac{1}{8}1^8-\frac{1}{5}1^5\Big)-\frac{1}{3}\Big(\frac{1}{8}0^8-\frac{1}{5}0^5\Big)\)

\(\displaystyle =\frac{1}{3}\Big(\frac{1}{8}-\frac{1}{5}\Big)-\frac{1}{3}\Big(0-0\Big)\)

\(\displaystyle =\frac{1}{3}\Big(\frac{5}{40}-\frac{8}{40}\Big)\)

\(\displaystyle =\frac{1}{3}\Big(-\frac{3}{40}\Big)=-\frac{3}{120}\)

Because there are no nested terms containing both \(\displaystyle x\) and \(\displaystyle y\), we can rewrite the integral as

\(\displaystyle \iint f(x,y)dxdy=\int3x^3dx \int \sqrt{y}dy+\int\sin(2x)dx \int \cos(4y)dy\)

This enables us to evaluate the double integral and the product of two independent single integrals.  From the integration rules from single-variable calculus, we should arrive at the result

\(\displaystyle \iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{12}\cos(2x)\sin(4y)+C\).

\(\displaystyle \int^2_1\int^2_0(x^3y)\:dxdy=\int^2_1(\frac{x^4y}{4}|^2_0)dy\int^2_14y\:dy =2y^2|^2_1=2(4-1)=2(3)=6\)

Because the x and y terms in the integrand are independent of one another, we can move them to their respective integrals:

\(\displaystyle \int dy \cdot \int x^2dx= \frac{x^3y}{3}+C\)

We used the following rules for integration:

\(\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}+C\), \(\displaystyle \int dx=x+C\)

First, you must evaluate the integral with respect to y (because of the notation \(\displaystyle dydx\)).

Using the rules of integration, this gets us 

\(\displaystyle 5x^4\frac{y^2}{2}dx\).

Evaluated from y=2 to y=3, we get 

\(\displaystyle 5x^4(\frac{9}{2}-\frac{4}{2})dx=\frac{25}{2}x^4dx\).

Integrating this with respect to x gets us \(\displaystyle \frac{5}{2}x^5\), and evaluating from x=0 to x=1, you get  \(\displaystyle \frac{5}{2}\).

First, you must evaluate the integral with respect to y and solving within the bounds.

In doing so, you get \(\displaystyle x^3y+\frac{y^3}{3}\) and you evaluate for y from 0 to 2.

This gets you 

\(\displaystyle \int_{1}^{2}(2x^3+\frac{8}{3})dx\).

This time evaluating the integral with respect to x gets you 

\(\displaystyle \frac{2x^4}{4}+\frac{8}{3}x\).

Evaluating for x from 1 to 2 gets you 

\(\displaystyle \frac{122}{12}\).

When solving double integrals, we compute the integral on the inside first.

\(\displaystyle \int_{0}^{7} \int_{0}^{5} 2x+7\,dxdy=\int_0^7x^2+7x|_{x=0}^{x=5}\,dy\)

\(\displaystyle =\int_0^75^2+7(5)-0^2+7(0)\,dy\)

\(\displaystyle =\int_0^760\,dy\)

\(\displaystyle =60y|_{y=0}^{y=7}\)

\(\displaystyle =420-0\)

\(\displaystyle =420\)

When solving double integrals, we compute the integral on the inside first.

\(\displaystyle \int_{0}^{6} \int_{0}^{4} 6y+8\,dxdy=\int_0^66xy+8x|_{x=0}^{x=4}\,dy\)

\(\displaystyle =\int_0^6 6(4)y+8(4)-6(0)y-8(0)\,dy\)

\(\displaystyle =\int_0^624y+32\,dy\)

\(\displaystyle =12y^2+32y|_{y=0}^{y=6}\)

\(\displaystyle =12(6)^2+32(6)-[12(0)^2+32(0)]\)

\(\displaystyle =624\)

When solving double integrals, we compute the integral on the inside first.

\(\displaystyle \int_{0}^{3} \int_{0}^{4} 3x+6\,dydx=\int_0^33xy+6y|_{y=0}^{y=4}\,dx\)

\(\displaystyle =\int_0^33x(4)+6(4)-[3x(0)+6(0)]\,dx\)

\(\displaystyle =\int_0^312x+24\,dx\)

\(\displaystyle =6x^2+24x|_{x=0}^{x=3}\)

\(\displaystyle =6(3)^2+24(3)-6(0)^2-24(0)\)

\(\displaystyle =126\)

First, you must evaluate the integral with respect to x. This gets you \(\displaystyle y^3\frac{x^2}{2}dy\) evaluated from \(\displaystyle x=0\) to \(\displaystyle x=2\). This becomes \(\displaystyle \int_{1}^{3}2y^3dy\). Solving this integral with respect to y gets you \(\displaystyle \frac{1}{2}y^4\). Evaluating from \(\displaystyle y=1\) to \(\displaystyle y=3\), you get \(\displaystyle \frac{1}{2}(\frac{3^4}{4}-\frac{1}{4})=10\).