# Definite Integrals

You might like to read Introduction to Integration first!

## Integration

Integration can be used to find areas, volumes, central points and many useful things. But it is often used to find the |
||

The area can be found by adding slices that And there are Rules of Integration that help us get the answer. |

## Notation

The symbol for "Integral" is a stylish "S" |

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:

### Example:

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

The **Indefinite**** Integral** is: **∫2x dx = x ^{2} + C**

- At x=1: ∫2x dx =
**1**^{2}+ C - At x=2: ∫2x dx =
**2**^{2}+ C

Subtract:

^{2}+ C) − (1

^{2}+ C)

^{2}+ C − 1

^{2}− C

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

In fact we can give the answer directly like this:

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

A = \frac{2+4}{2} × 1 = 3

Yes, it does have an area of 3.

(Yay!)

Let's try another example:

### 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):

**0.362...**

And another example to make an important point:

### Example:

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):

**0.460...**

That's better!

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

### Example:

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:

**−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):

### 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:

- 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:

From 1 to π/2:

**0.159...**

From π/2 to 3:

**−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.

## Continuous

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).

### Example:

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

## Properties

### 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:

## Summary

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