Fundamental Theorem of Calculus

[ralfsk8]

Homework Statement

All this information is in the attached file.

Homework Equations

All this information is in the attached file.

The Attempt at a Solution

What I tried to do was take the anti-derivative of the first equation and plug in the number 5. I'm not sure if that was the way to go so I then just plugged in 5 off the get-go and got 13. Not sure what the question is asking and I would appreciate some guidance.

Thanks

 

[Mark44]

Homework Statement

All this information is in the attached file.

Homework Equations

All this information is in the attached file.

The Attempt at a Solution

What I tried to do was take the anti-derivative of the first equation and plug in the number 5. I'm not sure if that was the way to go so I then just plugged in 5 off the get-go and got 13. Not sure what the question is asking and I would appreciate some guidance.

Thanks

If you have concluded that F(5) = 13, then no, that is incorrect. For that part you should NOT attempt to get the antiderivative.

 

[ralfsk8]

So the anti-derivative is only applicable to parts b and c?

 

[Mark44]

You should not attempt to get the antiderivative in any of the three parts. For part b, you'll need to use the Fundamental Theorem of Calculus. Note that there are two parts of this theorem.

 

[ralfsk8]

The book says to use the second part of the theorem but I'm not seeing how it would work in this question. Do I plug in the x in place of t?

 

[Mark44]

You're talking about part a, right? What definite integral is represented by F(5)?

 

[ralfsk8]

√5^2 + 144?

 

[Mark44]

√5^2 + 144?

No.
Show me what the definite integral looks like. If you quote this post you should be able to reuse this integral.

$$F(5) = \int_?^? \sqrt{t^2 + 144}~dt $$

 

[ralfsk8]

No.
Show me what the definite integral looks like. If you quote this post you should be able to reuse this integral.

$$F(5) = \int_5^5 \sqrt{t^2 + 144}~dt $$

Is it that?

 

[Mark44]

Yes. Now, what is the value of that integral? That is what F(5) equals. Don't overthink this.

 

[SammyS]

ralfsk8,

In general, what does the definite integral,

\displaystyle \int_{a}^{b}\, f(x)\,dx​

represent?

 

[ralfsk8]

Mark 44 - I would say either 13 or 0, but as you mentioned earlier, 13 isn't the answer.

SammyS - The area under the curve bounded by the x-axis of a graph and 2 vertical lines (a and b)

 

[Mark44]

What are the vertical lines for this integral?

 

[ralfsk8]

The vertical lines for this integral are 5 and x

 

[Mark44]

That's for F(x). You're trying to determine F(5). So what are the vertical lines now?

 

[STEMucator]

An easy way to remember the fundamental theorem is to consider :

Let :

F(x) = \displaystyle \int_{b(x)}^{a(x)}\, f(t)\,dt

Then :

\frac{d}{dx}F(x) = \frac{d}{dx} \displaystyle \int_{b(x)}^{a(x)}\, f(t)\,dt = f(a(x))a'(x) - f(b(x))b'(x)

It's simply a plug and play formula.

 

[ralfsk8]

Mark44 - The vertical lines are now 5 and 5. So we're not trying to calculate the area of anything if it's the same line?

Zondrina - Ah yes, that does seem familiar to me

 

[ralfsk8]

Mark44 - I would assume 5 to 5 which would make it the same line right?

Zondrina - Ah yes, that does seem familiar

 

[STEMucator]

Mark44 - I would assume 5 to 5 which would make it the same line right?

Zondrina - Ah yes, that does seem familiar

Also think about this F(5) thing, it's MUCH easier than it looks. Imagine you're integrating from one line a to another line b. What if those two lines are the same? Then there really is no area between them right?

A simple example is to consider :

\int_{z}^{z}xdx

Where z is ANY real number or function or anything really, but to keep it simple let z = 1,2,3... and notice no matter what you're going to get the same answer. Then come back to your F(5) question.

 

[Mark44]

Mark44 - I would assume 5 to 5 which would make it the same line right?

Right. So F(5) = ?

 

[ralfsk8]

Okay so I've narrowed it down to either 0 or 5...

 

[Mark44]

Okay so I've narrowed it down to either 0 or 5...

What's your reasoning for each of these?

 

[ralfsk8]

Well just like Zondrina said, if both of the vertical lines have the same value, then there shouldn't be area, deeming it 0.

As for the 5... I actually have no clue

 

[Mark44]

Well just like Zondrina said, if both of the vertical lines have the same value, then there shouldn't be area, deeming it 0.

As for the 5... I actually have no clue

OK, then what's your answer for F(5)?

 

[ralfsk8]

My final answer is 0.

 

[Mark44]

Right.

Now, what about parts b and c?

 

[ralfsk8]

Do I take the anti-derivative and plug in 5?

 

[STEMucator]

Do I take the anti-derivative and plug in 5?

No you use that plug and play i posted on the first page. Memorize it as it is your best friend to solve questions like these within a minute.

 

[Mark44]

Do I take the anti-derivative and plug in 5?

No. Here's what I said back in post #4.

You should not attempt to get the antiderivative in any of the three parts. For part b, you'll need to use the Fundamental Theorem of Calculus. Note that there are two parts of this theorem.

 

[Mark44]

No you use that plug and play i posted on the first page. Memorize it as it is your best friend to solve questions like these within a minute.

IMHO, memorizing that formula is NOT a good idea, especially if that memorization comes at the expense of understanding.

 

[ralfsk8]

So would the answer be:

x^2 + 144?

 

[Mark44]

For what? You should be writing an equation; i.e., something with = in it.

 

[ralfsk8]

I am completely lost :(

 

[Mark44]

You're given that $F(x) = \int_5^x \sqrt{t^2 + 144}~dt$

b) Find F'(5)

To do this part, first find F'(x), and then evaluate this derivative at x = 5. This is where the Fund. Thm. of Calculus comes into play.

 

[ralfsk8]

I'm comparing my answers with a similar problem in the book and there aren't any equations, only numerical answers.

 

[Mark44]

But your work should consist of equations, like this:

F'(x) = <some function of x that you work out>
F'(5) = <some number>

If you just throw something up in a post, I have no idea what you are doing.

 

[ralfsk8]

I understand the bit about the Fundamental Theorem of Calculus coming into play but I thought that you just substitute the bounds where the variables are. For example, the book gives us the example of:

Integral with lower bound of 1 and upper bound of x, (t^3)dt. The answer is given as merely x^3

 

[Mark44]

I understand the bit about the Fundamental Theorem of Calculus coming into play but I thought that you just substitute the bounds where the variables are. For example, the book gives us the example of:

Integral with lower bound of 1 and upper bound of x, (t^3)dt. The answer is given as merely x^3

OK, let me ask you: What does the answer you wrote represent?

 

[STEMucator]

IMHO, memorizing that formula is NOT a good idea, especially if that memorization comes at the expense of understanding.

Point taken I suppose haha, but grant me that I'm not going to write out and explain the whole proof of the theorem lol.

 

[Mark44]

Or, in other words, what question is x³ the answer to?

 

[ralfsk8]

Does it mean that the function is differentiable within those bounds?

 

[Mark44]

What function? Try to ask questions that are more precise.

 

[ralfsk8]

Okay I finally got the answer but only by comparing them to other online resources. I'm still not entirely sure on how to do the actual problem. I wouldn't mind discussing this further but if you guys need to go do other things, that's okay. Thanks for the help anyway.

 

[SammyS]

Okay I finally got the answer but only by comparing them to other online resources. I'm still not entirely sure on how to do the actual problem. I wouldn't mind discussing this further but if you guys need to go do other things, that's okay. Thanks for the help anyway.

ralfsk8,

Suppose that G(t) is an anti-derivative of \displaystyle \sqrt{t^2+144}\ .

We write that as \displaystyle G(t)=\int\,\sqrt{t^2+144}\ dt\ .

So if we have a definite integral such as \displaystyle \int_{a}^{b}\,\sqrt{t^2+144}\ dt\,, we can evaluate that as G(b) - G(a), according to the Fundamental Theorem of Calculus. Correct?

Now, in the case of the problem in this thread, we have:

\displaystyle F(x)=\int_{5}^{x}\,\sqrt{t^2+144}\ dt\,=G(x)-G(5)\ .​

Therefore, \displaystyle F&#039;(x)=G\,&#039;(x)\,, since G(5) is a constant.

But G(x) is the anti-derivative of \displaystyle \sqrt{x^2+144}\,, so that \displaystyle G\,&#039;(x)=\sqrt{x^2+144}\,. Correct?

Therefore, \displaystyle F&#039;(x)=\sqrt{x^2+144}\,.