Find the average value of the function over the indicated interval .

Question

Find the average value of the function  over the indicated interval  .

Quizplus · Accepted Answer

We can use the formula for average value of a function over an interval:

$avg = \frac{1}{b-a}\int_{a}^{b}f(x)dx$

In this case, we have $f(x) = 11\sin(x) + 12$, $a = 0$, and $b = \pi$. Plugging in these values, we get:

$avg = \frac{1}{\pi-0}\int_{0}^{\pi}(11\sin(x) + 12)dx$

$avg = \frac{1}{\pi}[(-11\cos(x) + 12x)]_{0}^{\pi}$

$avg = \frac{1}{\pi}[(-11\cos(\pi) + 12\pi) - (-11\cos(0) + 0)]$

$avg = \frac{1}{\pi}(-11\cos(\pi) + 12\pi + 11)$

Simplifying, we have:

$avg = \frac{1}{\pi}(11\pi + 11)$

$avg = \frac{11}{\pi}(\pi + 1)$

$avg \approx 13.2$

Rounded to the nearest whole number, the answer is C) 13.

Find the Average Value of the Function Over the Indicated