# Find the average temperature between $t=0$ and $t=24$ when $T(t) = 49+8t-(1/2)t^2$ degrees.

What was the average temperature during that period?

My initial thought was to take the derivative of the problem, plug in 24 for $t$ and solve. I was wrong. This is what I have

$T'=8-t=8-24=-16$ deg.

Should I have taken the integral instead of the derivative?

The average value of $f(t)$ over the interval $[a,b]$ is $$\frac{\int_a^b f(t)\,dt}{b-a}.$$ To see that derivative has not much to do with average value, suppose that $f(t)$ is the constant $K$ over our interval. Then the average value must be $K$. But $f'(t)=0$ for all $t$.
• One integral is $49t+4t^2-\frac{t^3}{6}$. Plug in $24$, take away the (easy) result of plugging in $0$, then divide by $24$. We get $49+4(24)-\frac{24^2}{6}$. – André Nicolas May 17 '13 at 23:09