For time $t\geq 0$, the height $h$ of an object suspended from a spring is given by $h(t)=5+8\cos\left(\dfrac{\pi t}{4}\right)$. What is the average height of the object from $t=0$ to $t=2$ ?

Summary Submit

The average value of a function is:

$$\frac{1}{b-a}\int\limits_a^b f(x)   dx$$ $$\frac{1}{2-0}\int\limits_0^2 5+8\cos\left(\frac{\pi t}{4}\right)   dt$$ $$= \frac{1}{2} \left[ 5t+\frac{32}{\pi}\sin\left(\frac{\pi t}{4}\right) \right]_0^2$$ $$= \frac{1}{2} \left[ \left(5(2)+\frac{32}{\pi}\sin\Big(\frac{\pi(2)}{4}\Big)\right)-\left(5(0)-\frac{32}{\pi}\sin\Big(\frac{\pi(0)}{4}\Big)\right) \right]$$ $$= \frac{1}{2}\left[ \left(10+\frac{32}{\pi}\right) - \left( 0-0\right) \right]$$ $$= \boxed{5+\frac{16}{\pi}}$$