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:

  integral area

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.

  integral area dx


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

  integral notation

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   definite integral
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:

definite integral y=2x from 1 to 2 as graph


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

definite integral 2x dx from 1 to 2


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

definite integral 2x dx from 1 to 2 = 2^2 - 1^2 = 3


area of y=2x from 1 to 2 equals 3

Check: with such a simple shape, let's also try calculating the area by geometry:

A = 2+42 × 1 = 3

Yes, it does have an area of 3.


Let's try another example:

definite integral y=cos(x) from 0.5 to 1 graph


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

definite integral cos(x) dx from 0.5 to 1

(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):

cos(x) dx
= sin(1) − sin(0.5)
 = 0.841... − 0.479...
 = 0.362...

And another example to make an important point:

definite integral y=sin(x) from 0 to 1 graph


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

definite integral sin(x) dx from 0 to 1


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):

sin(x) dx
= −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:

definite integral y=cos(x) from 1 to 3


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

definite integral cos(x) dx from 1 to 3

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:

cos(x) dx
= sin(3) − sin(1)
 = 0.141... − 0.841...
 = −0.700...

So there is more negative than positive parts, with the net result of −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 all area treated as positive (without the part below the axis being subtracted):

area y=cos(x) from 1 to 3 positive both above and below

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 all area is positive (imagine you had to paint it).

So now we have to do the parts separately:

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

From 1 to π/2:

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

From π/2 to 3:

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

That last one comes out negative, but we want it to be 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).

not continuous asymptote


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


Reversing the interval

definite integral negative property

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

definite integral a to b = negative of b to a


Interval of zero length

definite integral area zero

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

definite integral a to a = 0


Adding intervals

area a to b = a to c plus c to b

We can also add two adjacent intervals together:

definite integral a to b = a to c plus c to b


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