Block 2

Area Under Curves

What is to be learned?

• A practical use for integration!

y = x

42

Area?

2½ X 2X 2

=2

4

½ X 4X 4

= 8

Area = 8 – 2 = 6

y = x

42

Area?

2½ X 2 X 2

=2

4

½ X 4 X 4

= 8

Area = 8 – 2 = 6

2

4

= x2[ ]24

= 42 – 22

= 6

∫∫ x dx

2

2 2

Integration gives you area between graph and x axis

units2

y = 3x2

3

Area?

½ X2X2

0

3

= x32- 22

∫∫ 3x2 dx

2 2

= 27

= 33

[ ]3

0

– 0

units2

y = f(x)

b

Area?

½ X2X2 a

b

2- 22

∫∫ f(x) dx2 2

Area under a curveThe area between a curve and the x axis can be calculated using a definite integral

a

y = 4x3

2

Area?

½ X2X2

2- 22

2 2

1

1

2∫∫ 4x3 dx

= x41

= 15

= 24

[ ]2

– 14

units2

Find the area enclosed by y = 8x – 2x2 andthe x axis.Sketch – Find Roots

8x – 2x2 = 02x(4 – x) = 0x = 0 or x = 4

→ y = 0

40

4∫∫ (8x – 2x2)dx

= 211/3 units2= 4(4)2 – 2/3(4)3

= [4x2 – 2/3x3]40

– 0

Find the area enclosed by y = 9 – x2 andthe x axis.Sketch – Find Roots

9 – x2 = 0(3 – x)(3 + x) = 0x = 3 or x = -3

→ y = 0

3-3

3∫∫ (9 – x2)dx

= 36 units2= 9(3) – 1/3(3)3

= [9x – 1/3x3] 3-3

– ( 9(-3) – 1/3(-3)3 )

-3

Key Question

Try This

0

4∫∫ (x – 2)dx

0[ ]4x2 – 2x2

42 – 2(4) – 02

= 0!!!!!!!!!!!!!!!

y = x – 2

2

0

2∫∫ (x – 2)dx

2

4∫∫ (x – 2)dx

= 2= -2

Areas under x-axis will give a negative value!

y = 0

4

x = 4

Under the x Axis

Areas under the x axis will give a negative valueMay have to do 2 separate integrals

8

2

8∫∫ (2x – 8)dx

2

Need this point

y = 0

y = 2x – 8

2x – 8 = 0

x = 4

4

Area 1

2

4∫∫ (2x – 8)dx

2[ ]4x2 – 8x

= 42 – 8(4) –(22 – 8(2))= -16 + 12= -4Area 1 = 4units2

( )

x = 8

8

2

8∫∫ (2x – 8)dx

2

y = 2x – 8

4

Area 1

2

4∫∫ (2x – 8)dx

2[ ]4x2 – 8x

= 42 – 8(4) –(22 – 8(2))= -16 + 12= -4Area 1 = 4units2

( )

8

2

8∫∫ (2x – 8)dx

2

y = 2x – 8

4

Area 2

4

8∫∫ (2x – 8)dx

4[ ]8x2 – 8x

= 82 – 8(8) –(42 – 8(4))= 0 + 16= 16Area 2 = 16units2

Total area = 4 + 16 = 20 units2

( )