Fundamental Theorem of Calculus

**JonathanEyoon** · June 6th 2008, 11:51 AM

We just went over this in class and I am really confused by it. I know that there are two parts to the theorem and it is part 1 that I am confused by. Can someone please help me understand it? I can easily do the homework problems in the book by simply mimicking the examples but I'm not knowing the WHY which I think is far more important. Any help is much appreciated.

**Jhevon** · June 6th 2008, 11:54 AM

Originally Posted by JonathanEyoon

We just went over this in class and I am really confused by it. I know that there are two parts to the theorem and it is part 1 that I am confused by. Can someone please help me understand it? I can easily do the homework problems in the book by simply mimicking the examples but I'm not knowing the WHY which I think is far more important. Any help is much appreciated.

I've seen different parts labeled as part one and part two, which are you referring to specifically? how about a specific problem to work through?

The proof of the theorem should be in your texts, as well as all over the internet. you can look it up. that is really something you go into detail about with advanced calculus though...

**Mathstud28** · June 6th 2008, 11:55 AM

Originally Posted by JonathanEyoon

We just went over this in class and I am really confused by it. I know that there are two parts to the theorem and it is part 1 that I am confused by. Can someone please help me understand it? I can easily do the homework problems in the book by simply mimicking the examples but I'm not knowing the WHY which I think is far more important. Any help is much appreciated.

Is it this one

$\frac{d}{dx}\bigg[\int_a^{x}f(t)dt\bigg]=f(x)$

Or is it

$\int_a^{b}f(x)dx=F(b)-F(a)$

Where $F'(x)=f(x)$ ?

**JonathanEyoon** · June 6th 2008, 11:57 AM

MmMm... It is this one

Fundamental Theorems of Calculus -- from Wolfram MathWorld

where it is stating the second part. Sorry didn't know it gets referred as the second as well.

I don't really have a problem as such to work with it, it's just I'm not getting what it's saying and simply doing problems by mimicking the examples in my textbook.

For instance, i don't understand from beginning to end of what the second part is stating.

**JonathanEyoon** · June 6th 2008, 12:00 PM

The first one

**Mathstud28** · June 6th 2008, 12:01 PM

Originally Posted by JonathanEyoon

MmMm... It is this one

Fundamental Theorems of Calculus -- from Wolfram MathWorld

where it is stating the second part. Sorry didn't know it gets referred as the second as well.

I don't really have a problem as such to work with it, it's just I'm not getting what it's saying and simply doing problems by mimicking the examples in my textbook.

For instance, i don't understand from beginning to end of what the second part is stating.

Ok
Well its basically saying that once you find the antiderivative of a function, then the definite integral is just the antiderivative evaluated at the upper limit minus the the antiderivative evaluated at the lower limit (assuming hte integral is not improper, don't worry about this part, I jsut wanted a disclaimer)

**JonathanEyoon** · June 6th 2008, 12:03 PM

I understand the part of the FTC theorem where it is similar to the Evaluation and Net Change Theorem but I don't understand the other one where f(t) is involved.

**Mathstud28** · June 6th 2008, 12:08 PM

Originally Posted by JonathanEyoon

I understand the part of the FTC theorem where it is similar to the Evaluation and Net Change Theorem but I don't understand the other one where f(t) is involved.

That is the second fundamental theorem

Ok

The basic reasoning for it is this

we have that $\int_a^{x}f(t)dt$ where a is a constant and x is a variable

Then from that we have

$\int_a^{x}f(t)dt=F(x)-F(a)$ where $F'(t)=f(t)$

So now taking the deriative and remebering that snice $a$ is a constant $F(a)$ is also we get

$\frac{d}{dx}\bigg[F(x)-F(a)\bigg]=F'(x)$

and from the earlier assertion

$F'(x)=f(x)$

$\therefore\frac{d}{dx}\bigg[\int_a^{x}f(t)dt\bigg]=f(x)$

**JonathanEyoon** · June 6th 2008, 12:19 PM

Thanks alot fellas! Before you guys disappear, do you mind helping me work through 1 problem using that part of the theorem?

g(x) = from 0 = lower to x = upper (1 + t^(1/2))dt

so before I do anything, according to the theorem. This should = 1 + x^(1/2) right?

So basically we are taking the anti derivative of the function and integrating from 0 to x and then finding the derivative of whatever I get from there? Right?

**Mathstud28** · June 6th 2008, 12:23 PM

Originally Posted by JonathanEyoon

Thanks alot fellas! Before you guys disappear, do you mind helping me work through 1 problem using that part of the theorem?

g(x) = from 0 = lower to x = upper (1 + t^(1/2))dt

so before I do anything, according to the theorem. This should = 1 + x^(1/2) right?

So basically we are taking the anti derivative of the function and integrating from 0 to x and then finding the derivative of whatever I get from there? Right?

Your answer is correct, but remember the point of the second fundamental theorem is that you DO NOT have to integrate and differentiate, you can automatically write down the answer

**o_O** · June 6th 2008, 12:23 PM

Let $f(t) = 1 + t^{\frac{1}{2}}$ :

If you're asked to find g'(x) ...

$g(x) = \int_{0}^{x} f(t) dt$
$g'(x) = {\color{blue}\frac{d}{dx} \int_{0}^{x} f(t) dt}$
$g'(x) = {\color{blue}f(x)}$ (By the Fundamental Theorem of Calculus).

**JonathanEyoon** · June 6th 2008, 12:29 PM

Thanks alot everyone!

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