AREASUSING

INTEGRATION

We shall use the result that the area, A, bounded by a curve, y = f(x), the x axis

and the lines x = a, and x = b, is given by:

b

a

A y dx

Hence, the shaded area = (units squared)

Example 1: Find the area shown enclosed by the curvey = x2, the x-axis, and the lines x = 1 andx = 2.

I = 2

1

x2 dxx

3

3

2

1

=

2 3

3

=

1 3

3

–

73

=83

13

– 73=

Example 2: Find the area enclosed by the curve y = 3x2 – 6xand the x axis.

Firstly we need to find where the curve crosses the x-axis.

0 = 3x2 – 6x

x = 0 or 2

The graph looks like this:

= [ 8 – 12 ] – [ 0 ] = – 4

n.b. The integral is negative, this is because the area lies below the x axis. The actual area is 4 (units squared)

i.e. When y = 0;

0 = 3x(x – 2 )

Now, I = 3x2 – 6x dx2

0=

2

0

3x3

36x2

2– =

2

0

–x3 3x2

Since an area below the x axis givesa negative value, care must be takenwhen part of the required regionis above the x axis and part is below.

e.g.

Example 3: Find the area shown enclosed by the curve y = x ( x – 1), the x-axis, and the line x = 2 .

We need to calculate2 separate integrals:

1

0

23

1 23I

xx6

1

2

1

3

1

2

1

23

2 23I

xx

2

1

3

12

3

8

For the actual area, weadd the 2 answers,ignoring the negative:

1 6

5

6

1 Area 6

5

I1 = 1

0

x2 – x dx I2 = 2

1

x2 – x dxand,

i.e. Simultaneous equations are used to find the points of intersection.

y = f(x)

y = g(x)y

x

To find the area bounded bytwo curves, ( or a curve anda line ), consider the graphshown:

R

Then R = f(x) dxb

a

g(x) dxb

a

–

a b

Area bounded by two curves:

The required area, R, is the area under y = f(x) between x = a and x = b, minus the area under y = g(x) between x = a and x = b.

f(x) – g(x) dxb

a=RSince the limits are the same for both integrals,

they can be combined to a single integral:

Example 4:

R

x

yThe curve y = x2 – 4x + 5and the line y = 2x intersect

as shown. Find the area of the region Renclosed by the two graphs.

Firstly we need the two points of intersection, A and B.

A

B

x2 – 4x + 5 = 2x

x2 – 6x + 5 = 0

( x – 1 )( x – 5 ) = 0

The required area, R, is the area underthe line, between x = 1 and x = 5, minus the area under the curve betweenx = 1 and x = 5.

i.e. R = – x2 – 4x + 5 dx5

1

2x dx5

1

51

x = 1 or 5

Note: The 1st part of the integral 2x dx5

1i.e. ( the area under the line)

could be found by calculating the area of a trapezium.

6x5

1

– x2 – 5 dxR =

3x2 x 3

3–

– 5x

5

1

= ( 75 – 125 3

– 25 ) – ( 3 – 13

– 5 )

–2x dx5

1

x2 – 4x + 5 dx5

1R = We now have:

R =

32

3=

Since the limits are the same for both integrals, they can be combined to a single integral:

f(x) – g(x) dxb

a=Ri.e.

–2x5

1

( x2 – 4x + 5 ) dxR =

Summary of key points:

1. The area, A, bounded by a curve, y = f(x),the x axis and the lines x = a, and x = b, is given by:

b

a

dA xy

2. A region which lies below the x axis gives a negative result for the integral.

3. Care must be taken when part of the required region is above the x axis, and part is below.

This PowerPoint produced by R.Collins ; Updated Apr.2009

4. The area, R bounded by twofunctions which intersect at x = a, and x = b is given by:

f(x) – g(x) dxb

a=R

We shall now use the result that the area, A, bounded by a curve, x = f(y), the y axis and the lines y = a, and y = b, is given by:

b

a

A x dy

n.b. This is no longer on the syllabus.

Example 4: Find the area shown, bounded by the curve 3

xy

the y axis, and the lines y = 1, and y = 4.

Firstly we need the equation in theform x = f(y):

Square both sides:3

2 xy

So x = 3y2

dy3yA 4

1

2 4

1 3y = [ 64 ] – [ 1 ] = 63

dyxb

a

A Using,