# What is an integral?

1. May 2, 2012

### xeon123

I've this example of an integral.

$\int^{x}_{a} f(t)dt$

What t and dt means? Is there a relation between t and dt?

2. May 2, 2012

### HallsofIvy

Staff Emeritus
Since you are talking about an integral I assume you have some idea of what an integral is, either taking a course in Calculus or are reading a book on Calculus.

Every text book I have seen introduces the integral in terms of the Riemann sum in which you divide an area under the graph of y=f(x) into very thin rectangles- each having base $\Delta x$ and height $f(x_i^*)$ for $x_i^*$ being some x value within the base of the ith rectangle. The approximate area would be the sum of the areas of those rectangles: $\sum f(x_i^*)\Delta x$. One can show that the exact area is the limit of that as the base of those rectangles goes to 0. Then in the form $\int f(x)dx$ or $\int f(t)dt$ x and t are the independent variables in the functions f(x) and f(t), the limits of $f(x_i^*)$ and $f(t_i^*)$, and dx and dt are "infinitesmal" sections of the axis.

3. May 2, 2012

### xeon123

So what's the purpose of having dx in the expression? If I'm correct, dx means the derivative of x. And, in each point of x in the interval $\Delta x$, x will be a fixed value (is it a constant?), and the derivative of a constant is always 0.

4. May 2, 2012

### micromass

Staff Emeritus
The "dx" is just a notation. It doesn't mean anything. It's just used to denote which variable we integrate.

For example

$$\int_0^1 (t+x)dx$$

means that we integrate with respect to x. While

$$\int_0^1 (t+x)dt$$

means that we integrate with respect to t.

5. May 2, 2012

### xeon123

As far as I understand, in the first case we will only integrate x, and in the second case, t.

So, being f a function, f=(t+x)

In the first case, what happens to x? Can you integrate this example, please?

6. May 2, 2012

### micromass

Staff Emeritus
With the integral

$$\int (t+x)dx$$

we have that t is a constant and x is the integration variable, so

$$\int (t+x)dx = tx+\frac{x^2}{2}+C$$

While in

$$\int (t+x)dt$$

we treat x as a constant. Thus

$$\int (t+x)dx = \frac{t^2}{2}+tx+C$$

7. May 2, 2012

### xeon123

Thanks, now I got it.

8. May 3, 2012

### hunt_mat

I wrote some notes on integration, they should be available on the Math & Science Learning Materials section under notes on integration.