1. C1:Indefinite Integration
Learning Objective: to recognise integration as the reverse of differentiation

2. Starter:
Match the expressions of functions and their derivatives

3. Reversing the process of differentiation
To differentiate y = xn with respect to x we multiply by the power and reduce the power by one. xn
multiply by the power
reduce the power by 1
nxn-1
Suppose we are given the derivative = xn and asked to find y in terms of x.
Reversing the process of differentiation gives
divide by the power
increase the power by 1 xn
The process of finding a function given its derivative is called integration.

4. Reversing the process of differentiation
For example:
Adding 1 to the power and dividing by the new power gives:
= 2x3
This is not the complete solution, however, because if we differentiated y = 2x3 + 1,
or y = 2x3 – 3,
or y = 2x3 + any constant
we would also get
We therefore have to write y = 2x3 + c.

5. Reversing the process of differentiation
We can’t find the value of c without being given further information. It is called the constant of integration.
The integral of 6x2 with respect to x is written as:
Tell students that the integral symbol is derived from an elongated S. The reason for this will be explained when integration is used to find the area under a curve.
This rule will work for any positive, negative or fractional index except when the index is -1.

6. Examples
Integrate the following expressions with respect to x.
(a) 3x x + 3
(b) ½ x x6
(c) 2x3 − 3x-1/
(d) 3√x − x x3
(e) (x + 2) √x

9. Finding the constant of integration given a point
A curve y = f(x) passes through the point (2, 9).
find the equation of the curve.
Given that

10. The curve passes through the point (2, 9) and so we can substitute x = 2 and y = 9 into the equation of the curve to find the value of c.
y = 2x4 – 5x2 + c
9 = 2(2)4 – 5(2)2 + c
9 = 32 – 20 + c
9 = 12 + c
c = – 3
So the equation of the curve is y = 2x4 – 5x2 – 3.

11. Evaluating c :
Find the equation of the curve through (6, -18) such that dy/dx = x2 – 6x + 4.
The gradient of a curve at (x, y) is given by 1/x2 – 1/x3 and when x = 2, y = -1/8. Find the values of y when x = 4.
Find the equation of the curve through (-1, 5) for which dy/dx = 6(x2 – 1).