# Calc 2 problem involving integration

1. Mar 1, 2011

### bennwalton

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.

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

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

Last edited: Mar 1, 2011
2. Mar 1, 2011

### SammyS

Staff Emeritus
First of all, I assume you mean f(0) = 3.

How did you get 9/2 when integrating f '(x) from ‒3 to 0 ?
From ‒3 to ‒2, there is an area of 1/2 above the x-axis.
From ‒2 to 0, there is an area of 2, which is below the x-axis. That translated to an integral of ‒2.

That gives a result of ‒3/2.​

In general:

$$\int_a^bf'(x)\,dx=f(b)-f(a)$$

That's about all you need.

3. Mar 1, 2011

### bennwalton

Sorry about the notation mistake, having to hold down shift to do all my parentheses must have thrown me off; a mere typo.

Alas, I mixed up what I was doing when typing out my method.

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).

Does that make sense? I'm sorry my work is all kind of jumbled on my sheet, I wrote out my process for a different part of the problem but matched it with the answer I got for this one...

So in light of having done that, how would I find f(4)? The graph to the right of the origin is a semi circle with radius 2; x2 + y2 = 4. which would lead to y = sqrt (4 - x2). But that's as far as I know how to go

4. Mar 1, 2011

### SammyS

Staff Emeritus
Finding f(‒3):
Yes, that's one way to find f(‒3).

What I suggested went along with evaluating the integral graphically. There is more area below the x-axis than above on the interval [‒3, 0]. The area is ‒3/2.

Therefore, f(‒3) ‒ f(0) = ‒3/2 . Plug 3 in for f(0) and solve for f(‒3). Did you get f(‒3) = 3/2 ?​

Find f(4) in a similar way.

What is the area between f '(x) and the x-axis on the interval [0, 4] ?

It looks to me like it's the area of a 4×2 rectangle minus half the area of a circle of radius 2. (Of course this answer is negative since it's below the x-axis.)

Since f(x) is the anti-derivative of f '(x), the area is equal to f(4) ‒ f(0) .