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, b].

a and b (called limits, bounds or boundaries) are put at the bottom and top of the "S", like this:

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

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

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

Example: What is
definite integral 2x dx from 1 to 2

We are being asked for the Definite Integral, from 1 to 2, of 2x dx

First we need to find the Indefinite Integral.

Using the Rules of Integration we find that 2x dx = x2 + C

Now calculate that at 1, and 2:


(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 integral at x=1?

−cos(1) = −0.540...

What? It is negative? But it looks positive in the graph.

Well ... we made a mistake!

Because we also 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 regions, 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 value.


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 with a net result of −0.700....

So we have this important thing to remember:

f(x) dx  =  (Area above x axis) − (Area below x axis)


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


But sometimes we want all area treated as positive (without the part below the axis being subtracted).

In that case we must calculate the areas separately, like in this example:

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 we 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.


Area above − area below

The integral adds the area above the axis but subtracts the area below, for a "net value":

f(x) dx  =  (Area above x axis) − (Area below x axis)


Adding Functions

The integral of f+g equals the integral of f plus the integral of g:

f(x) + g(x) dx  =
f(x) dx  +
g(x) dx  


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.