Reviewing AP Calculus 1999 exam instantaneous rate problem

[disneychannel]

The graph of the function f, consisting of three line segments, is given below(#5)
"[URL
Number 5, collegeboard website[/URL]

a)compute g(4) and g(-2)
I have no idea how they got to that solution. I have been looking at it for a long time and still do not understand! Please help!

 

[fss]

g(x) is the integral of the function pictured. Did you calculate the integral representing g(4) and g(-2) ? There's nothing particularly difficult about those two integrals, assuming you know how to calculate an area.

 

[disneychannel]

I do not understand. I do not see how area is relevant in computing g(4) and g(-2). What I tried to do is make an equation for each line and find the antiderivative of it and plug in 4 and 1. I did get 3/2 for one of them, but not the other answers. I don't think that's what I am supposed to do. Please be more specific because I can not conceive your logic.

 

[Nebuchadnezza]

What is the antiderivative? The antiderivate is the AREA under the graph from a to b.
In your picture you can split the area under the graph into simple geometric shapes that are easy to calculate. I think you already know how to calculate the area of a triangle.

 

[fss]

I do not understand. I do not see how area is relevant in computing g(4) and g(-2). What I tried to do is make an equation for each line and find the antiderivative of it and plug in 4 and 1. I did get 3/2 for one of them, but not the other answers. I don't think that's what I am supposed to do. Please be more specific because I can not conceive your logic.

Perhaps you need to go back and review your notes then. You are given a function g(x) (which happens to be an integral) with the variable "x" as the upper limit of the integral. It should be quite obvious what the limits of integration are given the form of g(x) when finding g(4).

 

[disneychannel]

OKAY. Let me break this down a little more because maybe I am not explaining my situation well. A= integral from b to a[f(x)-g(x)] dx right? When I dealt with these kinds of problem, I had 2 equations(such a y=2x and y=-x). NOW, I am given a graph instead(ITS NOT IN MY NOTES). Having a boundary of 4 to 1 and breaking the segments into a geometric shapes DOES NOT give me 5/2. What is f(t)...f(4)=-1. how do you find the anitderivative of f(4)? PLEASE I need step-by-step information. Obviously, I do not get it and need thorough lubricated answers! Thank You!

 

[fss]

Honestly it sounds more like you're not trying/reading what we are telling you.

So, going back to the beginning: What is an integral? If you have a function plotted on a standard xy plane and are asked to shade in the integral of that function, what would you shade in? You are overthinking the problem. Nebuchadnezza already told you how to do it in post #4.

 

[paulfr]

For part a, you do not need an Antiderivative

[Note the Antiderivative is NOT the area, it is the function that allows calculating the area
using the 2nd Fundamental Theorem of Calculus; A = F(b) - F(a)]

The points are all integers so you can find the areas of the four regions by inspection and simple formulas of geometry.
g(4) = area from 1 to 4 = sum of areas from 1 to 2 + area 2 to 3 + area 3 to 4
g(4) = 1.5 + 1 - 1 = 1.5 ... note the minus sign for the last term as the area is under the axis

g(-2) = minus { 1/2 3 times 4 = 6 } = - 6
The minus is due to the direction of the limits; Int f(x) from a to b = minus Int f(x) from b to a

 

[Omega_Prime]

OKAY. Let me break this down a little more because maybe I am not explaining my situation well. A= integral from b to a[f(x)-g(x)] dx right? When I dealt with these kinds of problem, I had 2 equations(such a y=2x and y=-x). NOW, I am given a graph instead(ITS NOT IN MY NOTES). Having a boundary of 4 to 1 and breaking the segments into a geometric shapes DOES NOT give me 5/2. What is f(t)...f(4)=-1. how do you find the anitderivative of f(4)? PLEASE I need step-by-step information. Obviously, I do not get it and need thorough lubricated answers! Thank You!

I feel your frustration bro (been there). What happens when the function goes negative? (As far as integration is concerned)

 

[paulfr]

This statement in your Post # 3 may be your problem
"I do not understand. I do not see how area is relevant in computing g(4) and g(-2)."

This is a serious misunderstanding.
The function g(x) is the integral and the integral is the area under the integrand.
In this case the integrand is defined graphically.
But the function g(x) is still the area under the curve.

Also note you may have missed fundamentals
1/ Area below the X axis in considered negative
2/ Area computed from right to left is also considered negative
Integral f(x) from 1 to 3 = minus integral f(x) from 3 to 1

Hope this helps you move forward

Have you checked out www.khanacademy.org ??