I'm doing integration homework and a question that I am not sure how to approach popped up.
The velocity (in centimeters per second) of blood r cm from the central axis of an artery is given by $v(r) = k(R^2 − r^2)$, where k is a constant and R is the radius of the artery. Suppose that $k = 1200$ and $R = 0.3$. Find the average velocity v of the blood across a cross section of the artery.
Since I don't really know how to approach this problem and the question was on the section for integration I'll post what I did, but I know it isn't right. I'm just posting to show that I don't just want you all to do my homework.
So $v(r) = k(R^2 − r^2)$ so I plugged in the given $k$ and $R$. and I ended doing $1200\int .03^2-r^2dr$ and got $.0108-400r^3$ and solved for $r$ and hoped for the best, but it was futile. A step by step solution of this would be greatly appreciated. I know the answer is 72, I just do not know how to get to that answer. Thanks in advance.
So you want to find the average value of this function. In general, if you have an integrable function $f:[a, b] \rightarrow \mathbb{R}$, then the average value of that function is given by $\displaystyle \frac{1}{b-a}\int_a^b f(x) dx$. If you've been assigned this problem, you should be able to find this formula discussed somewhere in your textbook. (Maybe try the index?). Otherwise, you can find a quick proof of this formula by clicking here.
In our case, we have a function of $r$, and $r$ can vary anywhere from $0$ to $0.3$ (the velocity of the blood is changing depending only on the value of $r$). Therefore, setting up our integral, we have:
$$\text{Average Velocity} = \frac{1}{0.3}\int_0^{0.3}1200(0.3^2 - r^2)dr$$
And I'll leave that calculation for you. :)