Question

Evaluate each of the double integral in the exercise 37-54 as iterated integrals

$$\begin{gathered} \int {\int }_R(3-x+4y)dA \\ whereR=\left\{\right(x,y\left)\right|0\le x\le 1,-1\le y\le 3\} \end{gathered}$$

Short answer

The value of integral is 26units.

Step-by-step solution

Given information

We are given an integral as

$$\begin{gathered} \int {\int }_R(3-x+4y)dA \\ whereR=\left\{\right(x,y\left)\right|0\le x\le 1,-1\le y\le 3\} \end{gathered}$$

Step 2; Use iterated integrals

We get,

$$\begin{gathered} {\int }_{-1}^3{\int }_0^1(3-x+4y)dxdy \\ ={\int }_{-1}^3{{[3x-\frac{x^2}{2}+4xy]}^1}_{0}dy \\ ={\int }_{-1}^3[\frac{5}{2}+4y]dy \\ ={{[\frac{5}{2}y+2y^2]}^3}_{-1} \\ =26 \end{gathered}$$