The document provides steps to calculate the area between two curves by finding the intersection points using "y=y" and then integrating between those points. It works through multiple examples of finding the area between various curve combinations, including quadratic, cubic, and linear functions. The key steps are finding the intersection points using "y=y", setting up the integral as the top curve minus the bottom curve, and integrating between the bounds found from the intersection points.

1. Block 2
Area Between Curves

2. What is to be learned?
• How to find the area between two curves

3. So far
Can calculate areas of things like this
Quadratic Functions
Cubic Functions
All have
horizontal bases

4. y = f(x)
y = g(x)
a b
Area between curves
= ∫g(x) – ∫f(x)
= ∫ g(x) – f(x) dx
a
b
need points of
intersection
use y=y tactic
Area = ∫f(x)
a
b
Area = ∫g(x)
a
b

5. a b
= ∫ (-x2
+ 9x – 4) – (x2
+ x + 2) dx
1
3
y = x2
+ x + 2
y = -x2
+ 9x – 4
using y = y
x2
+ x + 2 = -x2
+ 9x – 4
2x2
– 8x + 6 = 0
2(x2
– 4x + 3) = 0
2(x – 3)(x – 1) = 0
x=3 or 1
1 3
top curve
Tidy up

6. Tidying Up
= ∫ (-x2
+ 9x – 4) – (x2
+ x + 2) dx
1
3
= ∫ (-x2
+ 9x – 4 – x2
– x – 2) dx
1
3
1
3
= ∫ (-2x2
+ 8x – 6) dx
= [-2
/3x3
+ 4x2
– 6x]
1
3
= -2
/3(3)3
+ 4(3)2
– 6(3) – [-2
/3(1)3
+ 4(1)2
– 6(1)]
= 0 – (-21
/3)
= 21
/3 units2

7. a b
Area = ∫ f(x) – g(x) dx
a
b
y = g(x)
y = f(x)
Tidy up before integrating
Area Between Curves
Find limits using y=y

8. = ∫ (-x2
+ 10x – 4) – (2x2
+ x + 2) dx
1
2
y = 2x2
+ x + 2
y = -x2
+ 10x – 4
using y = y
2x2
+ x + 2 = -x2
+ 10x – 4
3x2
– 9x + 6 = 0
3(x2
– 3x + 2) = 0
3(x – 2)(x – 1) = 0
x=2 or 1
1 2
top curve
Finding
limits

9. = ∫ (-x2
+ 10x – 4) – (2x2
+ x + 2) dx
1
2
= ∫ (-x2
+ 10x – 4 – 2x2
– x – 2) dx
1
2
1
2
= ∫ (-3x2
+ 9x – 6) dx
Now Integrate
Tidy Up
careful!

10. Find area enclosed by y = x2
and y = 3x – 2
y = y
x2
= 3x – 2
x2
– 3x + 2 = 0
(x – 2)(x – 1) = 0
x = 2 or x = 1
y = 3x – 2
2
= ∫ (3x – 2 – x2
) dx
y = x2
1
2
1
= [3
/2x2
– 2x – 1
/3x3
]
1
2
= 3
/2(2)2
– 2(2) – 1
/3(2)3
– [3
/2(1)2
– 2(1) – 1
/3(1)3
]
= -2
/3 – (-5
/6)
= 1
/6 units2
Key Question

11. Find area enclosed by y = 2x and
y = 4x – x2
y = y
2x = 4x – x2
x2
– 2x = 0
x(x – 2) = 0
x = 0 or x = 2
y = 2x
2
= ∫ (4x – x2
– 2x) dx
y = 4x – x2
0
2
= ∫ (2x – x2
) dx
0
2
= [x2
– 1
/3x3
]
0
2
= 22
– 1
/3(2)3
– 0
= 11
/3 units2