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 \(\displaystyle \frac{1}{2}(\frac{3^4}{4}-\frac{1}{4})=10\).

First, you must evaluate the integral with respect to z. Using the rules for integration, we get \(\displaystyle 3x^2z^2dx\) evaluated from \(\displaystyle z=3\) to \(\displaystyle z=6\). The result is \(\displaystyle \int_{0}^{1}54x^2dx\). This becomes \(\displaystyle 27x^3\), evaluated from \(\displaystyle x=0\) to \(\displaystyle x=1\). The final answer is \(\displaystyle 27\).

To evaluate the iterated integral, we start with the innermost integral, evaluated with respect to x:

\(\displaystyle \int_{y}^{2y}dx=x\left.\begin{matrix} 2y\\y \end{matrix}\right|=2y-y=y\)

The integral was found using the following rule:

\(\displaystyle \int dx=x+C\)

Now, we evaluate the last remaining integral, using our answer above from the previous integral as our integrand:

\(\displaystyle \int_{1}^{3}ydy=\frac{y^2}{2}\left.\begin{matrix} 3\\1 \end{matrix}\right|=\frac{9}{2}-\frac{1}{2}=4\)

The integral was found using the following rule:

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

To evaluate the double integral, compute the inside integral first.

\(\displaystyle \int_{0}^{1} \int_{0}^{2}x+y\,dx\,dy=\int_0^1 (\frac{1}{2}x^2+xy|_{x=0}^{x=2}dy\)

\(\displaystyle =\int_0^12+2y\,dy\)

\(\displaystyle =2y+y^2|_{y=0}^{y=1}\)

\(\displaystyle =2(1)+1^2-[2(0)+0^2]\)

\(\displaystyle =3\)

aTo evaluate the double integral, compute the inside integral first.

\(\displaystyle \int_{0}^{2} \int_{0}^{1}y-2x\,dx\,dy=\int_0^2 (xy-x^2|_{x=0}^{x=1}dy\)

\(\displaystyle =\int_0^2 y-1\,dy\)

\(\displaystyle =\frac{1}{2}y^2-y|_{y=0}^{y=2}\)

\(\displaystyle =\frac{1}{2}(2)^2-2-[\frac{1}{2}(0)^2-0]\)

\(\displaystyle =0\)

To evaluate the double integral, compute the inside integral first.

\(\displaystyle \int_{0}^{3} \int_{0}^{4}2x-2y\,dy\,dx=\int_0^3 2xy-y^2|_{y=0}^{y=4}dx\)

\(\displaystyle =\int_0^3 8x-16\,dx\)

\(\displaystyle =4x^2-16x|_{x=0}^{x=3}\)

\(\displaystyle =4(3)^2-16(3)-[4(0)^2-16(0)]\)

\(\displaystyle =-12\)

To perform the iterated integration, we must work from inside outwards. To start we perform the following integration:

\(\displaystyle \int_{0}^{5x}4ydy=2y^2 \Big|^{5x}_{0}= 2(5x)^2=50x^2\)

This becomes the integrand for the outermost integral:

\(\displaystyle \int_{0}^{1}50x^2dx=\frac{50x^3}{3}\Big|^{1}_{0}= \frac{50}{3}\)

To perform the iterated integral, we work from inside outwards.

The first integral we perform is

\(\displaystyle \int_{y}^{3y}2xdx=x^2\left.\begin{matrix} 3y\\y \end{matrix}\right|=9y^2-y^2=8y^2\)

This becomes the integrand for the outermost integral,

\(\displaystyle \int_{0}^{2}8y^2dy=\frac{8y^3}{3}\left.\begin{matrix} 2\\ 0 \end{matrix}\right|=\frac{64}{3}\)

To evaluate the iterated integral, we must work from inside outward.

The first integral we evaluate is

\(\displaystyle \int_{0}^{x^2}ydy= \frac{y^2}{2}\left.\begin{matrix} x^2\\0 \end{matrix}\right|= \frac{x^4}{2}\)

This becomes the integrand for the outermost integral.

The final integral we evaluate is

\(\displaystyle \int_{1}^{2}\frac{x^4}{2}dx= \frac{x^5}{10}\left.\begin{matrix} 2\\1 \end{matrix}\right|= \frac{32}{10}-\frac{1}{10}=\frac{31}{10}\)