Integrals Explained: What Does t and dt Mean?

[xeon123]

I've this example of an integral.

[itex]\int^{x}_{a} f(t)dt[/itex]

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

 

[HallsofIvy]

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 textbook 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 [itex]\Delta x[/itex] and height [itex]f(x_i^*)[/itex] for [itex]x_i^*[/itex] being some x value within the base of the ith rectangle. The approximate area would be the sum of the areas of those rectangles: [itex]\sum f(x_i^*)\Delta x[/itex]. 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 [itex]\int f(x)dx[/itex] or [itex]\int f(t)dt[/itex] x and t are the independent variables in the functions f(x) and f(t), the limits of [itex]f(x_i^*)[/itex] and [itex]f(t_i^*)[/itex], and dx and dt are "infinitesmal" sections of the axis.

 

[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 [itex]\Delta x[/itex], x will be a fixed value (is it a constant?), and the derivative of a constant is always 0.

 

[micromass]

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

For example

[tex]\int_0^1 (t+x)dx[/tex]

means that we integrate with respect to x. While

[tex]\int_0^1 (t+x)dt[/tex]

means that we integrate with respect to t.

 

[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?

 

[micromass]

With the integral

[tex]\int (t+x)dx[/tex]

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

[tex]\int (t+x)dx = tx+\frac{x^2}{2}+C[/tex]

While in

[tex]\int (t+x)dt[/tex]

we treat x as a constant. Thus

[tex]\int (t+x)dx = \frac{t^2}{2}+tx+C[/tex]

 

[xeon123]

Thanks, now I got it.

 

[hunt_mat]

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