C1:Indefinite Integration

Learning Objective: to recognise integration as the reverse of

differentiation

Starter:

Match the expressions of functions and their derivatives

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.

dy

dx

Reversing the process of differentiation gives

divide by the power increase the power by 1 xn

1

1

nx

n

The process of finding a function given its derivative is called integration.

Reversing the process of differentiation

For example: 2If = 6 find in terms of .dy

x y xdx

Adding 1 to the power and dividing by the new power gives:

36=

3y x

= 2x3

This is not the complete solution, however, because if we differentiated y = 2x3 + 1,

We therefore have to write y = 2x3 + c.

or y = 2x3 – 3,

or y = 2x3 + any constant

2= 6dy

xdx

we would also get

Reversing the process of differentiationWe 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:

2 36 = 2 +x dx x c

This rule will work for any positive, negative or fractional index except when the index is -1.

Examples

Integrate the following expressions with respect to x.

(a) 3x2 + 4x + 3

(b) ½ x2 + 7x6

(c) 2x3 − 3x-1/2 + 4

(d) 3√x − 5 + 4 x2 x3

(e) (x + 2)2

√x

Finding the constant of integration given a point

A curve y = f(x) passes through the point (2, 9). 38 10

dyx x

dx find the equation of the curve.Given that

3 3If = 8 10 then = (8 10 )dy

x x y x x dxdx

4 28 10

= +4 2

x xc

4 2= 2 5 +x x c

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.

Evaluating c :

1. Find the equation of the curve through (6, -18) such that dy/dx = x2 – 6x + 4.

2. 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.

3. Find the equation of the curve through (-1, 5) for which dy/dx = 6(x2 – 1).