Q. 50 - Calculus

Question

In Exercises 49–58, sketch the region determined by the iterated integral and then evaluate the integral. For some of these integrals, it may be helpful to reverse the order of integration.

${\int }_0^3{\int }_{y+2}^{2y-3}\left(x^2+3xy\right) dx dy$

Step-by-Step Solution

Answer

Value of the given integral is:$-\frac{1135}{8}$

Step 1. Given Information

An integral,${\int }_0^3{\int }_{y+2}^{2y-3}\left(x^2+3xy\right) dx dy$

Step 2. Sketching the region of integration

Region of integration is shaded in the graph below. 

 

Step 3. Evaluating the given iterated integral

Given integral,

$$\begin{gathered} {\int }_0^3{\int }_{y+2}^{2y-3}\left(x^2+3xy\right) dx dy \\ ={\int }_0^3{\overline{)\left(\frac{x^3}{3}+\frac{3x^{2}y}{2}\right)}}_{y+2}^{2y-3} dy \\ ={\int }_0^3\left(\frac{{(2y-3)}^3-{(y+2)}^3}{3}+\frac{3y\left[{(2y-3)}^2-{(y+2)}^2\right]}{2}\right) dy \\ ={\int }_0^3\left(\frac{(8y^3-27-36y^2+54y)-(y^3+8+6y^2+12y)}{3}+\frac{3y\left[(4y^2+9-12y)-(y^2+4+4y)\right]}{2}\right) dy \\ ={\int }_0^3\left(\frac{(7y^3-35-42y^2+42y)}{3}+\frac{3y(3y^2+5-16y)}{2}\right) dy \\ ={\int }_0^3\left(\frac{7y^3-42y^2+42y-35}{3}+\frac{9y^3-48y^2+15y}{2}\right) dy \\ ={\overline{)\left(\frac{{\displaystyle \frac{7y^4}{4}}-{\displaystyle \frac{42y^3}{3}}+{\displaystyle \frac{42y^2}{2}}-35y}{3}+\frac{{\displaystyle \frac{9y^4}{4}}-{\displaystyle \frac{48y^3}{3}}+{\displaystyle \frac{15y^2}{2}}}{2}\right)}}_0^3 \\ =\left[\frac{7\times 3^4}{12}-\frac{42\times 3^3}{9}+\frac{42\times 3^2}{6}-\frac{35\times 3}{3}+\frac{9\times 3^4}{8}-\frac{48\times 3^3}{6}+\frac{15\times 3^2}{4}\right]-\left[0\right] \\ =\frac{\left(567\times 6\right)-\left(1134\times 8\right)+\left(378\times 12\right)-\left(105\times 24\right)+\left(729\times 9\right)-\left(1296\times 12\right)+\left(135\times 18\right)}{72} \\ =\frac{3402-9072+4536-2520+6561-15552+2430}{72} \\ =-\frac{10215}{72} \\ =-\frac{1135}{8} \end{gathered}$$