Hey everyone! My name is Ben, I'm a senior in high school working on an BC Calc problem and am having some general questions about it.(adsbygoogle = window.adsbygoogle || []).push({});

The question reads:

Let f be a function defined on the closed interval [-3,4] with f(0) = 3. The graph of f', the derivative of f, consists of one line segment and a semi circle, as shown above.

Then they provide this image:

http://t0.gstatic.com/images?q=tbn:ANd9GcS5nCFcatsgGUIOnZXhKAObqzlF7qkiuW5MOX13BQcePo0e-N3jGQ&t=1

Letter D reads:

Find f(-3) and f(4). Show the work that leads to your answers.

To find f(-3), I found an equation for the line of f'(x) on x: [-3,0]. I found the equation to be y' = -x - 2. I then recovered the original function on that same interval, and found y = -x2/2 - 2x + C. I used the point (0,3) to find C because of the f(0) = 3 and found that f(x) = -x2 - 2x + 3 (all of this only on the interval [-3,0]. I then plugged in -3 into this recovered equation to find f(-3) = 9/2.

For whatever reason though, the second question baffled me. I tried to integrate geometrically again (as from 0 to 4 it's a semicircle of r=2 cut out of a rectangle of dimensions 2x4) and came up with the answer 8-2π (that's supposed to be a pi symbol...).

I guess my primary question is, do I add to that the integral of the other side? Would f(4) be the integral from [-3,0] + [0,4], or simply the integral from [0,4]? Do I use the equation of a circle to do something similar; do I have to integrate that?

Thanks so much! It's a wonder that a site like this exists :P

**Physics Forums - The Fusion of Science and Community**

# Calc 2 problem involving integration

Have something to add?

- Similar discussions for: Calc 2 problem involving integration

Loading...

**Physics Forums - The Fusion of Science and Community**