# Integration is the inverse process of differentiation

1. Sep 5, 2008

### JinM

Hi everyone,
I know that integration is the inverse process of differentiation, and that the definite integral is defined as:
$\int_{a}^{b} f(x) dx = \lim_{n \to \infty} \sum^{n}_{i = 1} f(x_i) \Delta x$
assuming that the integrand is defined over the interval [a,b].

My question is: Why is it that the antiderivatives give the area under the curve? For example:

$\int \frac{1}{1+x^2} dx = \arctan{x} + C$

How does arctan give the area under curve of the integrand between a and b?

Thanks and much appreciated,
JinM

2. Sep 5, 2008

### morphism

Re: Integral

Strictly speaking, the antiderivative doesn't give you the area under the curve. It's the definite integral that does. The reason for this lies in the way the integral is constructed.

3. Sep 5, 2008

### JinM

Re: Integral

But you know the point I'm trying to express. Let F(x) be the antiderivative of f(x). To find the area under the curve of f(x) between a and b, we evaluate: F(b)-F(a). Why?

I want to see how evaluating the antiderivative gives you the area under the curve.

4. Sep 5, 2008

### morphism

Re: Integral

The key to seeing that is a good understanding of how the Riemann integral is constructed, via partitions and rectangles, etc. and an appreciation of what the mean value theorem really says.

If you let {a=x_0 < x_1 < ... < x_n=b} be a partition of [a,b], then

$$F(b) - F(a) = F(x_n) - F(x_0) = \sum_{i=1}^n F(x_i) - F(x_{i-1}).$$

Now apply the mean value theorem to get a c_i in each partition [xi-1, xi] such that

$$F(x_i) - F(x_{i-1}) = F'(c_i) (x_i - x_{i-1}) = f(c_i) \Delta x_i.$$

If we plug this back into the original equation, we get that

$$F(b) - F(a) = \sum_{i=1}^n f(c_i) \Delta x_i.$$

And 'letting n tend to infinity' will now yield

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

5. Sep 5, 2008

### JinM

Re: Integral

Thanks for your help so far.

I'm sorry, but I may not have been clear with my question; I'm not a good expositor. :l

I understand how the Riemann integral is constructed. I want to know why the inverse operation of differentiation gives you the function that yields the area under the curve.

6. Sep 5, 2008

### morphism

Re: Integral

But.. I just showed you why!

7. Sep 5, 2008

### neg_ion13

Re: Integral

An anti-derivative is not the area under the curve. An integral is. That being said, the area under the curve is the geometric interpretation. The best way to see it is to look at the definition of the integral itself. Sf(x)dx means sum of f(x) which is the y value, times dx which is the delta x segments on the x axis. This means, height (y value) * base (dx) which is the area of a rectangle. the S is the sum of all the rectangles. The fundamental theorem of calculus is the algebraic interpretation, the "getting back from the derivative".

8. Sep 6, 2008

### HallsofIvy

Staff Emeritus
Re: Integral

The fundamental theorem of calculus:
If dF/dx= f(x), then $$\int f(x)dx= F(b)- F(a) and If $F(x)= \int_a^x f(t)dt$, then dF/dx= f(x) One way to show that is this: If y= f(x)> 0, between x= a and x= b, then we can identify (or define) [tex]\int_a^x f(t)dt[/itex] as "the area of the region bounded by the graphs of y= f(x), y= 0, x= a, and x= b". Given any x0, $F(x_0)= \int_a^{x_0} f(t)dt$ is the area under the curve from a to x0 and, for any h> 0, $F(x+h)= \int_a^{x_0+ h}f(t)dt$ is the area under the curve from a to x0+ h. The difference, F(x0+ h)- F(x0) is the area under the curve from x0 to x0+ h. Suppose f(x*) is the minimum value for f on that interval, x0 to x0+ h. Then clearly the rectangle with base h and height f(x*) is completely contained in that region. Suppose f(x*) is the maximum value for f on that interval, then the rectangle with base h and height f(x*) contains that region. In terms of area, $hf(x_*)\le \int_{x_0}^{x_0+ h} f(x) dx\le hf(x^*)$. Dividing by h, [tex]f(x_*)\le \frac{\int_{x_0}^{x_0+ h} f(t)dt}{h}\le f(x^*)$$
but the area between x0 and x0+ h is just the area from a to x0 minus the area from a to x0. That is:
$$f(x_*)\le \frac{F(x+h)- F(x_0)}{h}\le f(x^*)$$

Now take the limit a h-> 0. Since x0+ h goes to x0, and both x* and x*[/sub] lie between them, both x* and x* go to x0 so
[tex]f(x_0)\le \frac{dF}{dx}(x_0)\le f(x_0)[/itex]
That is, dF/dx evaluated at x=x0, is just f(x0).