# Understanding integration

• Oct 20th 2009, 04:22 PM
gralla55
Understanding integration
While I do understand that

lim(n->(infinity) epsilon(i=1,n)f(xi)*(delta)x

should give an area under f(x), I fail to understand why this equation equals F(b)-F(a). I mean, why is the antiderivate interpreted as the area under a curve? I've looked in my math-book and searched wikipedia, but they only write that the closed integral is defined as the above formula, with no further explanation. This has been driving me crazy lately, if anyone could explain or post a link to an explanation I would greatly appreciate it.
• Oct 20th 2009, 04:26 PM
Quote:

Originally Posted by gralla55
While I do understand that

lim(n->(infinity) epsilon(i=1,n)f(xi)*(delta)x

should give an area under f(x), I fail to understand why this equation equals F(b)-F(a). I mean, why is the antiderivate interpreted as the area under a curve? I've looked in my math-book and searched wikipedia, but they only write that the closed integral is defined as the above formula, with no further explanation. This has been driving me crazy lately, if anyone could explain or post a link to an explanation I would greatly appreciate it.

Are you familiar with the fundamental theorem of calculus? There are two parts to the theorem, and one of them will help you understand.
• Oct 20th 2009, 04:38 PM
gralla55
Quote:

Are you familiar with the fundamental theorem of calculus? There are two parts to the theorem, and one of them will help you understand.
I'm looking at it now. If I understand it correctly, the first part only says that the derivate of the integral equals the original function, which is fairly obvious.

As for the second part... I'm not really sure what it says.
• Oct 20th 2009, 08:51 PM
calum
The integral is defined as the area under a curve f(x) between a and b.

It is the Fundamental Theorem of Calculus that proves that this area is equal to difference in value between a and b of the original function of which f(x) is the derivative. This is the second part of the theorem and a the proof is shown on the Wikipedia page.
• Oct 20th 2009, 09:14 PM
Bruno J.
Quote:

Originally Posted by gralla55
I'm looking at it now. If I understand it correctly, the first part only says that the derivate of the integral equals the original function, which is fairly obvious.

As for the second part... I'm not really sure what it says.

Obvious if you've defined the integral of $\displaystyle f$ as the function whose derivative is $\displaystyle f$. Defining the integral as the (signed) area under the curve of $\displaystyle f$, then it's a theorem.
• Oct 20th 2009, 09:15 PM
Bruno J.
Quote:

Originally Posted by calum
The integral is defined as the area under a curve f(x) between a and b.

It is the Fundamental Theorem of Calculus that proves that this area is equal to difference in value between a and b of the original function of which f(x) is the derivative. This is the second part of the theorem and a the proof is shown on the Wikipedia page.

Of any function whose derivative is $\displaystyle f$. There are infinitely many of them but any one will do.
• Oct 21st 2009, 12:26 AM
gralla55
Quote:

Obvious if you've defined the integral of http://www.mathhelpforum.com/math-he...1929cce7-1.gif as the function whose derivative is http://www.mathhelpforum.com/math-he...1929cce7-1.gif. Defining the integral as the (signed) area under the curve of http://www.mathhelpforum.com/math-he...1929cce7-1.gif, then it's a theorem.
Sure, that makes sense. But what I'm wondering is "why" the formula for the area of the curve happens to be the opposite of that of differentiating.
• Oct 21st 2009, 01:22 AM
calum
Quote:

Originally Posted by gralla55
Sure, that makes sense. But what I'm wondering is "why" the formula for the area of the curve happens to be the opposite of that of differentiating.

Try and visualise this (it might help if you draw diagrams):

- You have some curve f(x) and you want to find the area under a small section of the curve at point (a, f(a))

- Then, because this area is small, it can be approximated by the rectangle $\displaystyle f(a)\times \Delta x$

- Now imagine tht you have a curve g(x), and you want to find the small change in the value of the function when you have a small change in x, $\displaystyle \Delta x$, at the point (a, g(a)).

-Since the change in x is small, the small change in the function value will approximately equal the gradient at the point a multiplied by the small change in x (approximating the curve with a line).

- That is $\displaystyle \Delta g(x) = g'(a) \times \Delta x$

-Now we can see that $\displaystyle \Delta g(x) =$ area under f(x) at the point a only if $\displaystyle f(a) = g'(a)$. That is, the area under f(x) is only equal to the change in g(x) if g(x) is the anti-derivative of f(x).

What I have written is not mathematically rigorous, but I hope it will help you visualise the relationship.
• Oct 21st 2009, 07:57 AM
Bruno J.
Quote:

Originally Posted by gralla55
Sure, that makes sense. But what I'm wondering is "why" the formula for the area of the curve happens to be the opposite of that of differentiating.

For that you would need to look at the proof of the first part of the Fundamental Theorem. So it's not so obvious after all eh! (Wink)

The above poster sketched the proof for you. Good luck!
• Oct 21st 2009, 12:44 PM
gralla55
Now I'm getting somewhere!
Quote:

Try and visualise this (it might help if you draw diagrams):

- You have some curve f(x) and you want to find the area under a small section of the curve at point (a, f(a))

- Then, because this area is small, it can be approximated by the rectangle http://www.mathhelpforum.com/math-he...870284ca-1.gif

- Now imagine tht you have a curve g(x), and you want to find the small change in the value of the function when you have a small change in x, http://www.mathhelpforum.com/math-he...96519319-1.gif, at the point (a, g(a)).

-Since the change in x is small, the small change in the function value will approximately equal the gradient at the point a multiplied by the small change in x (approximating the curve with a line).

- That is http://www.mathhelpforum.com/math-he...48e51037-1.gif

-Now we can see that http://www.mathhelpforum.com/math-he...f98fab33-1.gif area under f(x) at the point a only if http://www.mathhelpforum.com/math-he...22e86976-1.gif. That is, the area under f(x) is only equal to the change in g(x) if g(x) is the anti-derivative of f(x).
The first two points I understand, I even made a drawing. If you could elaborate a little on points 3-5 I it would help me alot. Thanks!
• Oct 21st 2009, 07:38 PM
Bruno J.
Let $\displaystyle A(x)$ be the area under the curve of $\displaystyle f(x)$ between some fixed number $\displaystyle a$ and $\displaystyle x$. Then for a small increase from $\displaystyle x$ to $\displaystyle x+\delta$, the area increases approximately by the area of a rectangle of height $\displaystyle f(x)$ and base $\displaystyle \delta$. Hence

$\displaystyle A(x+\delta) \approx A(x) + f(x) \delta$

or

$\displaystyle \frac{A(x+\delta)-A(x)}{\delta} \approx f(x)$.

As $\displaystyle \delta$ goes to 0, the left side converges to the derivative of $\displaystyle A(x)$, and the right side to $\displaystyle f(x)$. The argument may be made precise but the above is essentially the proof of the first part of the FTC.
• Oct 21st 2009, 08:26 PM
calum
I have attached a handwritten page explaining points 3-5 in more detail, as it would take too long with Latex! Hope it helps.
• Oct 25th 2009, 02:34 PM
gralla55
Wow, thanks alot for taking the time to write all that by hand! I'm starting to get the general idea, but I'll have to read it over some more to truly "get" it. I actually found a different proof in my old high-school math-book. Maybe I'll post it here later.