# Integration

Before going too far down this notes sheet, ensure you're on top form for everything differentiation... else this will be harder than it needs to be!

**Integration is the opposite of differentiation**, and can be used to find the initial function form its derivative. For example, to find f(x) from f'(x), you integrate. It is also commonly used to find the area between a curve and the x-axis.

## Integrating xⁿ

When differentiating xⁿ, you multiply by the power and then reduce the power by one. The reverse of this is to **add one to the power and divide by this**:

If f'(x) = xⁿ, then f(x) = (xⁿ⁺¹) / (n+1)

If there is a coefficient of x, you do the same but:

If f'(x) =kxⁿ, then f(x) =k(xⁿ⁺¹) / (n+1)

Note the difference between differentiating and integrating xⁿ:

Just like when differentiating polynomials (functions with multiple terms),

**integrate one term at a time.**

## Definite & Indefinite Integrals

Integration is often noted using **∫ () dx**, where the elongated S tells you to integrate the function in the brackets while the dx tells you to do so with respect to x. **It is important to get this notation right.**

When there are no limits with the ∫, the integral is called

**indefinite**, because it can only**give a function**with an unknown y-intercept,*c*.

If there are limits attached to the ∫,

*a*and*b*, the integral is**definite**, as it**produces a value**for a defined interval [a, b].

## Indefinite Integral Functions

Unless you are given more information, **indefinite integrals give an unspecific function**. This means they produce a function but do not locate it - as in it could intersect anywhere on the y-axis. This is because when you differentiate a function, the y-intercept is not an x-term, so disappears. Therefore, when integrating the derivative, you have no idea where the y-intercept is, so we note it as an **unknown, c.**

You always need to add ' +c' to unspecific integral functions, else it is not complete.

### Finding the y-intercept, *c*

However, when you are given more information than just what to integrate, for example a coordinate that the function passes through, you can find the value of *c*, the y-intercept to find the **particular integral**:

## Areas under Graphs

When given a **definite integral**, you integrate the function, and then **evaluate** it by substituting in the limits and subtracting the lower from the upper:

∫[a, b] f'(x) dx = f(b) - f(a)

For example, to evaluate **4x+3**:

While differentiation is used to find the gradient of a function,** integration is used to give the area** enclosed by the graph and the x-axis and the limits.

Therefore, the example above is actually how to calculate the area under the line y = 4x + 3 between x=1 and x-3:

Evaluating definite integrals gives the area enclosed by the function, the limits and the x-axis.

### Areas under the x-axis

If the graph is below the x-axis, the value will be negative.

Therefore, when there is a root in an interval, you need to find the root and split it up into two separate definite integrals and evaluate them separately.

Find the x-coordinate of the root, r

Evaluate two integrals, one on the interval [r, b] and one over [a, r]

Add the positive integral to the modulus of the negative integral

### Area under Parametric Curves

This requires the standard function for integrating 1/x - see below

It is not necessary to find the Cartesian form of a curve given as parametric to integrate it. Instead, we can use the chain rule to write *dx* as *dx/dt*, and integrate with respect to the parameter,* t*:

Ifxandyare given as functions oft, ∫y dx= ∫y dx/dt dt

**It is important to remember to change the limits into***t*, as you are integrating with respect to*t*.

## Area Between Graphs

When trying to find the area enclosed by two or more graphs, it is easiest to split it up into the separate parts from each graph and the add or subtract accordingly.

### Areas between curves & lines

To find the area enclosed between a curve and a straight line, you need to find the area between each and the x-axis and then subtract one from the other. Often, you will have to use equations of triangle and trapezium area:

Area of a triangle = ½ ab sinC

Area of a trapezium = ½ (a+b)h

### Areas between curves

To find the area between two curves, you use integration to calculate the area between each curve and the x-axis, and then subtract one from another.

## Integrating Standard Functions

We have covered how to integrate xⁿ, but there are nine more standard functions you need to know:

∫ xⁿ dx = (xⁿ⁺¹) / (n+1) +c

∫eˣ dx =eˣ +c

∫ 1/x dx = ln|x| +c

∫ cos(x) dx = sin(x) +c

∫ sin(x) dx = -cos(x) +c

∫ tan(x) dx = ln|sec(x)| +c

∫ sec²(x) dx = tan(x) +c

∫ cosec(x) cot(x) dx = -cosec(x) +c

∫ cosec²(x) dx = -cot(x) +c

∫ sec(x) tan(x) dx = sec(x) +c

∫ f'(ax+b) dx = f(ax+b) /a+c

### Vital things to know

The last standard function

**only works for functions in the form f(ax+b)**and**not for trigonometric functions**like cos(ax+b).**Only the trigonometric functions listed here can be integrated.**For other functions, such as to integrate tan(x), you need to use trigonometric identities (see notes sheet on trigonometry) to get them into a form that can be integrated.Often, to get a function into a format that can be integrated you need to split it into

**partial fractions**. Then, you can use the standard result for 1/x, or the reverse chain rule (see below)

## Reverse Chain Rule

The **reverse chain rule** can be used when functions can be written in the form ** k f'(x) / f(x)**. The numerator must be an exact multiple of the derivative of the denominator for this to apply.

For example: