Area Under Curves

1. Block 2
Area Under Curves

2. What is to be learned?
• A practical use for integration!

3. y = x
42
Area?
2
½ X 2X 2
=2
4
½ X 4X 4
= 8
Area = 8 – 2 = 6

4. y = x
42
Area?
2
½ X 2 X 2
=2
4
½ X 4 X 4
= 8
Area = 8 – 2 = 6
2
4
= x2
[ ] 2
4
= 42
– 22
= 6
∫∫ x dx
2
2 2
Integration gives
you area between
graph and x axis
units2

5. y = 3x2
3
Area?
½ X2X2
0
3
= x3
2
- 22
∫∫ 3x2
dx
2 2
= 27
= 33
[ ]
3
0
– 0
units2

6. y = f(x)
b
Area?
½ X2X2 a
b
2
- 22
∫∫ f(x) dx
2 2
Area under a curve
The area between a curve and the x
axis can be calculated using a definite integral
a

7. y = 4x3
2
Area?
½ X2X2
2
- 22
2 2
1
1
2
∫∫ 4x3
dx
= x4
1
= 15
= 24
[ ]
2
– 14
units2

8. Find the area enclosed by y = 8x – 2x2
and
the x axis.
Sketch – Find Roots
8x – 2x2
= 0
2x(4 – x) = 0
x = 0 or x = 4
→ y = 0
4
0
4
∫∫ (8x – 2x2
)dx
= 211
/3 units2
= 4(4)2
– 2
/3(4)3
= [4x2
– 2
/3x3
]
4
0
– 0

9. Find the area enclosed by y = 9 – x2
and
the x axis.
Sketch – Find Roots
9 – x2
= 0
(3 – x)(3 + x) = 0
x = 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

10. Try This
0
4
∫∫ (x – 2)dx
0
[ ]
4
x2
– 2x
2
42
– 2(4) – 0
2
= 0!!!!!!!!!!!!!!!

11. 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

12. Under the x Axis
Areas under the x axis will give a negative
value
May have to do 2 separate integrals

13. 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
[ ]
4
x2
– 8x
= 42
– 8(4) –(22
– 8(2))
= -16 + 12
= -4
Area 1 = 4units2
( )
x = 8

14. 8
2
8
∫∫ (2x – 8)dx
2
y = 2x – 8
4
Area 1
2
4
∫∫ (2x – 8)dx
2
[ ]
4
x2
– 8x
= 42
– 8(4) –(22
– 8(2))
= -16 + 12
= -4
Area 1 = 4units2
( )

15. 8
2
8
∫∫ (2x – 8)dx
2
y = 2x – 8
4
Area 2
4
8
∫∫ (2x – 8)dx
4
[ ]
8
x2
– 8x
= 82
– 8(8) –(42
– 8(4))
= 0 + 16
= 16
Area 2 = 16units2
Total area = 4 + 16 = 20 units2
( )