QUESTION 2 Find the average value of the function f(x,y) = 20 - 2y over the rectangle R = [0,3] x [0,5]. A f_ave = 0 B f_ave = 75 C f_ave = 15 D f_ave = 3 E None of these

QUESTION 2 Find the average value of the function f(x,y) = 20 – 2y over the rectangle R = [0,3] x [0,5]. A f_ave = 0 B f_ave = 75 C f_ave = 15 D f_ave = 3 E None of these

Answer: 15

Explanation: Subject: Mathematics — specifically multivariable calculus (double integrals). Use the average-value formula for a function over a region:

\[ f_{\text{avg}}=\frac{1}{\text{Area}(R)}\iint_R f(x,y)\,dA. \]

Here \(\text{Area}(R)=3\cdot5=15\). Compute the double integral by iterated integration (Fubini’s Theorem) of \(f(x,y)=20-2y\) over \(x\in[0,3],\,y\in[0,5]\), then divide by 15.

Steps:

  1. Write the average-value expression:

\[ f_{\text{avg}}=\frac{1}{15}\int_{x=0}^{3}\int_{y=0}^{5}(20-2y)\,dy\,dx. \]

  1. Compute the inner integral (with respect to \(y\)):

\[ \int_{0}^{5}(20-2y)\,dy = \left[20y - y^2\right]_{0}^{5} = 100 - 25 = 75. \]

  1. Integrate with respect to \(x\):

\[ \int_{0}^{3}75\,dx = 75\cdot 3 = 225. \]

  1. Divide by the area:

\[ f_{\text{avg}}=\frac{225}{15}=15. \]

Therefore the average value is 15 (choice C).

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