# Thread: Working with the Fundamental Theorem of Calculus

1. ## Working with the Fundamental Theorem of Calculus

I have a problem that involves a triangle.
We are supposed to find the area - of course we can use 1/2 bh.
The entire question is like this:

Consider the graph of the continuous function f in the figure and let F(x)=0-x f(t) dt.
Assume that the graph consists of a line segment from (0,-2) to (2,2) and two quarter circles of radius 2 (see screenshot of graph below).
(Problems 1-3n these pages are based on the function y=f(t) shown on the graph (see screenshot below)
Question #1 Evaluate F(1). I used the 1/2 bh formula and found that the area was 1. This answer is correct.
Question#2 Evaluate F'(1). Use Fundamental theorem of Calculus, part 1. The answer was 0
Question #3 Evaluate F''(1). Use Fundamental Theorem of Calculus. The answer is 2

I found the area was -1. This was the answer to the question "Evaluate F(1)"
After that I was asked the following:

What is F'(1). I assume this would be the derivative of -1, which would be zero? The answer is zero. Not sure if this is the reason
The last question is "Evaluate F''(1). Apparently the answer is 2, but I have no idea why.
I was told this: F''(x)=f'(x) and F''(1)=f'(1)=2. Can someone help out? The graph is below:

calc3.jpg

2. Originally Posted by calc67x
I have a problem that involves a triangle.
We are supposed to find the area - of course we can use 1/2 bh.
I found the area was -1. This was the answer to the question "Evaluate F(1)"
After that I was asked the following:

What is F'(1). I assume this would be the derivative of -1, which would be zero? The answer is zero. Not sure if this is the reason
The last question is "Evaluate F''(1). Apparently the answer is 2, but I have no idea why.
I was told this: F''(x)=f'(x) and F''(1)=f'(1)=2. Can someone help out? The graph is below:

calc3.jpg
F'(1) is kind of a notation abuse - but a very common one. It REALLY means F'(x) evaluated at x = 1.
F'(1) is very clearly the slope of the line segment. Calculate from the endpoints.
F"(1) is very clearly 0. It's a line segment.

3. Originally Posted by calc67x
I have a problem that involves a triangle.
We are supposed to find the area - of course we can use 1/2 bh.
I found the area was -1. This was the answer to the question "Evaluate F(1)"
After that I was asked the following:

What is F'(1). I assume this would be the derivative of -1, which would be zero? The answer is zero. Not sure if this is the reason
The last question is "Evaluate F''(1). Apparently the answer is 2, but I have no idea why.
I was told this: F''(x)=f'(x) and F''(1)=f'(1)=2. Can someone help out? The graph is below:

calc3.jpg
Please state the problem exactly as given, or show the whole thing. They have defined f(t), but you are talking about F(x), so you have left something out. (It looks like you may also be thinking they are the same thing.)

I suspect that it says somewhere that, whereas the graph is for f(t), they told you that F(x) is the integral of f(t) from t=0 to x. If so, then you are correct about F(1), which would be a (signed) area. But probably that is not how the question was phrased.

Then, F'(x) = f(x), so F'(1) means f(1), which is 0. It does not mean the derivative of -1, because F(x) is not -1; that's only F(1).

Then, F"(1) is the derivative of F'(x) at x=1, which is the derivative of f(t) at t=1, namely 2.

4. Dr. Peterson:
Thanks for trying to help.

I am still trying to understand your sentence:
Then, F'(x) = f(x), so F'(1) means f(1), which is 0. It does not mean the derivative of -1, because F(x) is not -1; that's only F(1).

Why does F'(1) result in zero? I am not understanding how to interpret this or F''(1)=2.
What does F'(1) actually mean? Sorry, I am just not quite getting it. I am able to get the area of the region, which is -1 and then I can't get any further in my understanding.

Originally Posted by Dr.Peterson
Please state the problem exactly as given, or show the whole thing. They have defined f(t), but you are talking about F(x), so you have left something out. (It looks like you may also be thinking they are the same thing.)

I suspect that it says somewhere that, whereas the graph is for f(t), they told you that F(x) is the integral of f(t) from t=0 to x. If so, then you are correct about F(1), which would be a (signed) area. But probably that is not how the question was phrased.

Then, F'(x) = f(x), so F'(1) means f(1), which is 0. It does not mean the derivative of -1, because F(x) is not -1; that's only F(1).

Then, F"(1) is the derivative of F'(x) at x=1, which is the derivative of f(t) at t=1, namely 2.

5. Originally Posted by calc67x
The entire question is like this:

Consider the graph of the continuous function f in the figure and let F(x)=0-x f(t) dt.
Assume that the graph consists of a line segment from (0,-2) to (2,2) and two quarter circles of radius 2 (see screenshot of graph below).
(Problems 1-3n these pages are based on the function y=f(t) shown on the graph (see screenshot below)
Question #1 Evaluate F(1). I used the 1/2 bh formula and found that the area was 1. This answer is correct.
Question#2 Evaluate F'(1). Use Fundamental theorem of Calculus, part 1. The answer was 0
Question #3 Evaluate F''(1). Use Fundamental Theorem of Calculus. The answer is 2
Presumably you meant F(x)=∫0x f(t) dt.

Originally Posted by calc67x
I am still trying to understand your sentence:
Then, F'(x) = f(x), so F'(1) means f(1), which is 0. It does not mean the derivative of -1, because F(x) is not -1; that's only F(1).

Why does F'(1) result in zero? I am not understanding how to interpret this or F''(1)=2.
What does F'(1) actually mean? Sorry, I am just not quite getting it. I am able to get the area of the region, which is -1 and then I can't get any further in my understanding.
It sounds like what you aren't getting is the Fundamental Theorem of Calculus (FTC). Do you feel that you understood what you were taught about it?

The FTC says that if F(x)=∫0x f(t) dt, then F'(x) = f(x). Therefore, F'(1) = f(1), right? And, from the graph, that is 0.

Then, differentiating F'(x) = f(x), we have F"(x) = f'(x).

One way to think about the FTC and what F'(x) means is that if you add a little bit (∆x) to x, you are extending the area representing F(x) a little to the right. The increase is approximately a rectangle with width ∆x and height f(x); so the rate of change of F(x) is f(x)∆x/∆x = f(x).

6. ## Thanks - I think I'm getting it

Dr. Peterson:

Ok, I think I get it now! So the function of 1 is actually 0, it can be seen from the graph.
And if we check 0, the graph will point to 2. I believe I was trying to actually take a derivative of these numbers and at this point the language of the FTC is difficult for me, so I am confused.

Thanks for the help!

Originally Posted by Dr.Peterson
Presumably you meant F(x)=∫0x f(t) dt.

It sounds like what you aren't getting is the Fundamental Theorem of Calculus (FTC). Do you feel that you understood what you were taught about it?

The FTC says that if F(x)=∫0x f(t) dt, then F'(x) = f(x). Therefore, F'(1) = f(1), right? And, from the graph, that is 0.

Then, differentiating F'(x) = f(x), we have F"(x) = f'(x).

One way to think about the FTC and what F'(x) means is that if you add a little bit (∆x) to x, you are extending the area representing F(x) a little to the right. The increase is approximately a rectangle with width ∆x and height f(x); so the rate of change of F(x) is f(x)∆x/∆x = f(x).

7. Originally Posted by calc67x
Dr. Peterson:

Ok, I think I get it now! So the function of 1 is actually 0, it can be seen from the graph.
And if we check 0, the graph will point to 2. I believe I was trying to actually take a derivative of these numbers and at this point the language of the FTC is difficult for me, so I am confused.

Thanks for the help!
Right. It does take some time to fully grasp that you can let the theorem do all the work for you, when you've been putting so much effort into learning to do derivatives yourself!

In effect, you are given the graph of the derivative (of F); so the second derivative of F is the slope of the graph.

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