evaluate the double integral by first identif...
evaluate the double integral by first identifying it as the volume of a solid. ∬r(4 - 2y)da, r = 0, 1 × 0, 1
Answer
# Answer:
3
# Explanation:
## Step1: Set up the double - integral
$$\iint_R(4 - 2y)dA=\int_{0}^{1}\int_{0}^{1}(4 - 2y)dydx$$
## Step2: Integrate with respect to y
First, integrate $\int_{0}^{1}(4 - 2y)dy$. Using the power - rule $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int_{0}^{1}(4 - 2y)dy=\left[4y-y^{2}\right]_{0}^{1}=4\times1 - 1^{2}-(4\times0 - 0^{2})=4 - 1=3$.
## Step3: Integrate the result with respect to x
Now, we have $\int_{0}^{1}3dx$. Since $\int_{0}^{1}3dx=3x\big|_{0}^{1}=3\times1-3\times0 = 3$.