for #1 - 4, find the average values of the following functions over their given intervals. 1. f(x)=√(36…

Answer:

$\frac{3\pi}{2}$

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(x)$ over the interval $[a,b]$ is given by $\bar{y}=\frac{1}{b - a}\int_{a}^{b}f(x)dx$. Here, $a=-6$, $b = 6$, and $f(x)=\sqrt{36 - x^{2}}$, so $\bar{y}=\frac{1}{6-(-6)}\int_{-6}^{6}\sqrt{36 - x^{2}}dx=\frac{1}{12}\int_{-6}^{6}\sqrt{36 - x^{2}}dx$.

Step2: Recognize the geometric shape

The function $y = \sqrt{36 - x^{2}}$ can be rewritten as $x^{2}+y^{2}=36$ ($y\geq0$), which represents the upper - half of a circle with radius $r = 6$.

Step3: Calculate the integral

The integral $\int_{-6}^{6}\sqrt{36 - x^{2}}dx$ is the area of the upper - half of the circle with radius $r = 6$. The area of a full circle is $A=\pi r^{2}$, so the area of the upper - half of the circle is $A=\frac{1}{2}\pi r^{2}$. Substituting $r = 6$, we get $\int_{-6}^{6}\sqrt{36 - x^{2}}dx=\frac{1}{2}\pi(6)^{2}=18\pi$.

Step4: Find the average value

Substitute the value of the integral into the average - value formula: $\bar{y}=\frac{1}{12}\times18\pi=\frac{3\pi}{2}$.