
#1
Mar111, 06:38 PM

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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.
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:A...Po0eN3jGQ&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 82π (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 



#2
Mar111, 08:59 PM

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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 xaxis.In general: [tex]\int_a^bf'(x)\,dx=f(b)f(a)[/tex] That's about all you need. 



#3
Mar111, 09:36 PM

P: 2

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 = x^{2}/2  2x + C. I used the point (0,3) to find C because of the f(0) = 3 and found that f(x) = x^{2}  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; x^{2} + y^{2} = 4. which would lead to y = sqrt (4  x^{2}). But that's as far as I know how to go 



#4
Mar111, 11:10 PM

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Calc 2 problem involving integration
Finding f(‒3):
Yes, that's one way to find f(‒3).Find f(4) in a similar way. What is the area between f '(x) and the xaxis 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 xaxis.) Since f(x) is the antiderivative of f '(x), the area is equal to f(4) ‒ f(0) . 


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