Definite Integrals

You might like to read Introduction to Integration first!


Integration can be used to find areas, volumes, central points and many useful things. But it is often used to find the area under the graph of a function like this:


The area can be found by adding slices that approach zero in width:

And there are Rules of Integration that help us get the answer.



The symbol for "Integral" is a stylish "S"
(for "Sum", the idea of summing slices):


After the Integral Symbol we put the function we want to find the integral of (called the Integrand),

and then finish with dx to mean the slices go in the x direction (and approach zero in width).

Definite Integral

A Definite Integral has start and end values: in other words there is an interval (a to b).

The values are put at the bottom and top of the "S", like this:

Indefinite Integral
(no specific values)
  Definite Integral
(from a to b)

We can find the Definite Integral by calculating the Indefinite Integral at points a and b, then subtracting:


The Definite Integral, from 1 to 2, of 2x dx:


The Indefinite Integral is: 2x dx = x2 + C

  • At x=1: 2x dx = 12 + C
  • At x=2: 2x dx = 22 + C


(22 + C) − (12 + C)
22 + C − 12 − C
4 − 1 + C − C = 3

And "C" gets cancelled out ... so with Definite Integrals we can ignore C.

In fact we can give the answer directly like this:


We can check that, by calculating the area of the shape:

Yes, it has an area of 3.


Let's try another example:


The Definite Integral, from 0.5 to 1.0, of cos(x) dx:

(Note: x must be in radians)


The Indefinite Integral is: cos(x) dx = sin(x) + C

We can ignore C when we do the subtraction (as we saw above):

  = sin(1) − sin(0.5)
    = 0.841... − 0.479...
    = 0.362...

And another example to make an important point:


The Definite Integral, from 0 to 1, of sin(x) dx:


The Indefinite Integral is: sin(x) dx = −cos(x) + C

Since we are going from 0, can we just calculate the area at x=1?

−cos(1) = −0.540...

What? The Area at x=1 is negative? No, we need to subtract the integral at x=0. We shouldn't assume that it is zero.

So let us do it properly, subtracting one from the other (and C gets cancelled so we don't need to show it):

  = −cos(1) − (−cos(0))
    = −0.540... − (−1)
    = 0.460...

That's better!

But we can have negative areas, when the curve is below the axis:


The Definite Integral, from 1 to 3, of cos(x) dx:

Notice that some of it is positive, and some negative.
The definite integral will work out the net area.


The Indefinite Integral is:cos(x) dx = sin(x) + C

So let us do the calculations:

  = sin(3) − sin(1)
    = 0.141... − 0.841...
    = −0.700...

Try integrating cos(x) with different start and end values to see for yourself how positives and negatives work.

But sometimes you want the actual area (without the part below being subtracted):

Example: What is the area between y = cos(x) and the x-axis from x = 1 to x = 3?

This is like the example we just did, but area is positive (imagine you had to paint it).

So now we have to do the parts separately:

  • One for the area above the x-axis
  • One for the area below the x-axis

The curve crosses the x-axis at x = π/2 so we have:

cos(x) dx = sin(π/2) − sin(1)
      = 1 − 0.841...
= 0.159...
cos(x) dx = sin(3) − sin(π/2)
      = 0.141... − 1
  = −0.859...

That last one comes out negative, but we want positive, so:

Total area = 0.159... + 0.859... = 1.018...

This is very different from the answer in the previous example.


Oh yes, the function we are integrating must be Continuous between a and b: no holes, jumps or vertical asymptotes (where the function heads up/down towards infinity).


A vertical asymptote between a and b affects the definite integral.


Reversing the interval

Reversing the direction of the interval gives the negative of the original direction.


Interval of zero length

When the interval starts and ends at the same place, the result is zero:


Adding intervals

We can also add two adjacent intervals together:


The Definite Integral between a and b is the Indefinite Integral at b minus the Indefinite Integral at a.