Question

Earlier in this section, we showed that we could use Fubini’s theorem to evaluate the integral $\int {\int }_R{x}^{2}ydA$and we showed that ${\int }_2^5{\int }_1^3{x}^{2}ydxdy=91$Now evaluate the double integral by evaluating the iterated integral${\int }_1^3{\int }_2^5{x}^{2}ydydx$.

Short answer

The solution is,

${\int }_1^3{\int }_2^5{x}^{2}ydydx=91$.

Step-by-step solution

Step 1. Given Information.

The iterated integral is${\int }_1^3{\int }_2^5{x}^{2}ydydx$.

Step 2. Evaluating the integral.

We integrate with respect to $x$.

$$\begin{gathered} {\int }_1^3{\int }_2^5{x}^{2}ydydx \\ ={\int }_1^3\left({\int }_2^5{x}^{2}ydy\right)dx \\ ={\int }_1^3\left({\int }_2^5{x}^2\left(ydy\right)\right)dx \\ ={\int }_1^3{\left[x^2\frac{y^{1+1}}{1+1}\right]}_{y=2}^{y=5}dx \\ ={\int }_1^3{\left[\frac{1}{2}{x}^2{y}^2\right]}_{y=2}^{y=5}dx \\ ={\int }_1^3\left(\frac{1}{2}{x}^2{5}^2-\frac{1}{2}{x}^2{2}^2\right)dx \\ ={\int }_1^3\left(\frac{25}{2}{x}^2-2x^2\right)dx \\ ={\int }_1^3\left(\frac{21}{2}{x}^2\right)dx \\ =\frac{21}{2}{\int }_1^3{x}^{2}dx \\ =\frac{21}{2}{\left[\frac{x^{2+1}}{2+1}\right]}_{x=1}^{x=3} \\ =\frac{21}{2}{\left[\frac{x^3}{3}\right]}_{x=1}^{x=3} \\ =\frac{21}{2}\left[\frac{3^3}{3}-\frac{1^3}{3}\right] \\ =\frac{21}{2}\left[\frac{27}{3}-\frac{1}{3}\right] \\ =\frac{21}{2}\times \frac{26}{3} \\ =7\times 13 \\ =91 \end{gathered}$$