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# Integration

If we were to look at a function on a graph, the integral of would describe the area underneath the function. To represent the integral of , we would write , with telling us we are integrating with respect to x. (This is called the differential).

Conceptually, we think of integration as being the inverse of differentiation. This means that to find an integral, we can think of having to 'undo' the process of differentiation. When we have 'undone' the integration, we call this result the antiderivative.

## Indefinite integrals

The purpose of an indefinite integration is to find the antiderivative. The antiderivative is given as a function and doesn't tell us directly the area under the function. If we want to check whether we have the correct antiderivative, we can differentiate the antiderivative, and we should arrive back at the original function. If our original function is , we often denote as the antiderivative of .

When we find an indefinite integral, it is important that we add a constant of integration, meaning that if we were to find , we would give our answer as . This reflects that this antiderivative function could have any constant and still differentiated to the original function.

Find

differentiates to , so that is our antiderivative. However, in full, our answer is , as we must include this constant of integration.

## Definite integration

A definite integral is one with limits, so we could view this as the area under a function between two points, say point a and point b. For a function , we would write this as . This can be visualized as

Visualisation of a definite integral

The way to visualize this is to split the area under the function into n equal strips between a and b. This means we have the width of each strip, . We then take the height of each strip as , with the point at some point in strip i. This is shown below.

Visualisation of a definite integral

The area of the strips at this point is given as . To find the value of the integral, we need to use an infinite number of strips to cover the inside of the curve fully. This means as we take the limit, we get .

In practice, this becomes easier as we find the antiderivative (without the integration constant) and then evaluate it at the two limits, taking the bottom limit away from the top limit.

Find

The first step is to find the antiderivative of . This means we need to think of a function that differentiates to . Thinking about this, we get to . Now we know the antiderivative; we need to evaluate this at the limits.

### Properties of the integral

• For constants and , and functions , then

• for

## Integration methods

Not all antiderivatives can be found easily by inspection. Here, we can use an integration method instead to allow us to find the antiderivative.

### Integration by parts

By the product rule (as seen in differentiation), for two functions and then

If we integrate both sides with respect to x, we get:

which then simplifies to -

We rearrange this to -

This is now our formula for integration by parts, and we will demonstrate this through an example.

Use integration by parts to find .

We are going to let and . We now look to find and differentiating, we find and integrate to find . This means that .

We can now evaluate this last integral to give .

Note did we have included the integration constant here. We could have included this earlier; however, we can combine them all into one here.

### Integration by substitution

There is also an option to use substitution to simplify an integral. Here we change the variable we integrate with respect to. In the case of a definite integral, the limits also need to be changed using the substitution. We must also change the integrand. This is best demonstrated by an example. Trying to choose the right substitution takes time. However, it becomes easier.

Use substitution to find

Let us take , which means that . We can rearrange this to get .

Now substituting this in, we get

.

### Parametric integration

We can also be given a function parametrically and be expected to integrate this. Suppose we are given that

and , then the integral of the curve defined by these functions is given as . We can think of this as the 's canceling to give , which is what we'd expect in a normal integral.

Suppose we are given a curve defined by , with t ranging from 0 to 1, and we want to find the area under this curve w hen , so our integral is given as . We can evaluate this to get

## Examples of integration

While integrating, there are lots of handy rules to know. Over the next section, we will have a look at some of these.

### Integration of polynomials

From the derivative of a polynomial, you should know that . For an integral, we can reverse this to get . This rule will become second nature the more integrals that you do.

Integrate with respect to x.

### Integrating

The above formula for polynomials will not work for . So, let's look at this a different way:

. Let , then so .

Filling this in, we get -

Note we put x in a modulus function to ensure that the logarithm input is valid. We can extend this further. By making a suitable substitution, we can show -

.

### Integrating trigonometric functions

Like everything we have seen so far, we can treat integration as the inverse of differentiation, and this continues with trigonometric functions. We may have to use substitutions to solve these, and we can also introduce trigonometric functions as a substitution.

Find .

, Now let and then . This means that

Use a trigonometric substitution to find .

Let so , and .

Substituting this in, we get

.

As . So,

## Integration - Key takeaways

• Integration is the inverse of differentiation.

• A definite integral is bounded by limits and is evaluated there.

• An indefinite integral is an antiderivative and an integration constant.

• Integration by parts is defined as

• For a polynomial,

Integration is the process of finding the antiderivative of a function.

An example of integration would be Integration of  3x^2dx=x^3+C

Integration by parts is when we have two functions multiplied together, and then we have an integral of this. The equation for this is given by

Int of u(x)v'(x)dx=u(x)v(x)- Int of u'(x)v(x)dx

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