1. Finding the area under a curve
using definite integral
y y
x = f(y)
y = f(x)
d
A
A c
x x
0 a b 0
d
b
∫ f ( x)dx definite integral ∫ f ( y)dy
c
a
area of the shaded region

2. Finding the area under a curve
using definite integral
y
y = f(x)
(6,9)
2
x
6
(a) Shaded the region of ∫ f ( x)dx
0 6
(b) Hence, find the value of ∫ ydx + ∫ xdy
0

3. Finding the area under a curve
using definite integral
y
y = f(x)
(6,9)
9
2 ∫ xdy
2
x 6 9
6 ∫ ydx + ∫ xdy = (6 × 9) = 54
∫
0
f ( x)dx 0 2

4. Finding the area under a curve
using definite integral
y
y = 4x2
Find the area of the shaded region
x
0 4

5. Finding the area under a curve
using definite integral
area of the shaded region
Find the area of the 4
shaded region = ∫ ydx
y 0
4
= ∫ 4 x 2 dx
y = 4x2 0
4
 4x 3
= 
 3 0
 4(4) 3 4(0) 3 
x = − 
0 4 3 3 

1
= 85
3

6. Finding the area under a curve
using definite integral
y
y2 = 3x
y=4
Find the area of the
y=2
shaded region
x
0

7. Finding the area under a curve
using definite integral
Find the area of the area of shaded region
shaded region 4
y = ∫ xdy
2
y = 3x
2
4
y2
y=4 = ∫ dy
2
3
4
y=2  y 3
= 
 3(3)  2
0
x  (4) 3 (2) 3 
= − 
 9 9 
2
=6
9

8. Finding the area under a curve
using definite integral
Find the area of the shaded region
y
y = 2x y
y = 2x
P(3,6)
4 P(3,6)
y = 5x – x2
y = 5x – x2
A B
Q(5,0)
x
0 Q(5,0)
x
0 3 5
A is the area under a straight line
B is the area under a curve

9. Finding the area under a curve
using definite integral
Find the area of the shaded region
Area A = area of triangle
y = 1/2 x 3 x 6
y = 2x
= 9 unit2
5
4 P(3,6)
Area B = ∫ ydx
y = 5x – x2
3
5
A B = ∫ (5 x − x 2 )dx
3
Q(5,0)
x 5
0 3 5  5x 2
x 
3
= − 
 2 3 3
 5(5) 2 53   5(3) 2 33 
Area of shaded region =
 2 − 3 − 2 − 3 
  
= Area A + Area B    
1
1 1 =7
= 9 + 7 = 16 3
3 3

10. Finding the area under a curve
using definite integral
Find the area of the shaded region
y
y = 2x
6 P(5,5)
y = 5x – x2
x = 6y – y2
A
x
0

11. Finding the area under a curve
using definite integral
Find the area of the shaded region
Area A = area of triangle
y = 1/2 x 5 x 5
y = 2x
= 12.5 unit2
6
6 B
P(5,5) Area B = ∫ xdy
5 y = 5x – x2
5
A x = 6y – y2 6
A = ∫ (6 y − y 2 )dy
5
x 6
0 6y 2
y 
3
= − 
 2 3 5
 6(6) 2 63   6(5) 2 53 
Area of shaded region =
 2 − 3 − 2 − 3 
  
= area A + area B    
2
2 1 =2
= 12.5 + 2 = 15 3
3 6

12. Finding the area under a curve
using definite integral
Find the coordinate of P and Q.
Hence, find the area of the shaded region
y
Coordinates P and Q:.
y = 6x + 4
y = 6x + 4 …… ( 1 )
y = – 3x2 + 10x + 8 …….. ( 2 )
Q per (1) = per (2)
y = – 3x2 + 10x + 8 6x + 4 = – 3x2 + 10x + 8
3x2 - 10x - 8 + 6x + 4 = 0
P
x 3x2 - 4x - 4 = 0
0
(3x + 2)(x – 2) = 0
subtitute x = -2/3 into eq (1) x = -2/3 atau x = 2
y = 6(-2/3 ) + 4 = 0
subtitute x = 2 into eq (1)
y = 6(2 ) + 4 = 16
Koordinat P(-2/3 , 0) dan Q(2, 16)

13. Finding the area under a curve
using definite integral
Find the coordinate of P and Q.
Hence, find the area of the shaded region
y
y = 6x + 4
Q
y = – 3x2 + 10x + 8
P
x
0 2

14. Finding the area under a curve
Area of the shaded region using definite integral
y = 6x + 4
y
Q
y = – 3x2 + 10x + 8
P
0 2
x
Area under curve
2
Area under straight line
2 Area = ∫ ydx
Area = ∫ ydx 0
2
= ∫ (−3 x 2 + 10 x + 8)dx
2
0
= ∫ (6 x + 4)dx 0 2
0 2  − 3 x 3 10 x 2 
 6x2  = + + 8x
= + 4 x  3 2 0
 2 0
 6(2)2
  − 3(2) 3 10(2) 2 
= + 4(2)  − 0 =
 3 + + 8(2)  − 0

 2   2 
 
=28 unit2.
=20 unit . 2

15. Finding the area under a curve
Area of shaded region using definite integral
y = 6x + 4
y
y
Q
Q
y = – 3x2 + 10x + 8
P
0 2
x
P
0 x
y = 6x + 4 Area under curve
y 2
Area under straight line
2 Area = ∫ ydx
Area = ∫ ydx
Q 0
2
= ∫ (−3 x 2 + 10 x + 8)dx
2
0
= ∫ (6 x + 4)dx 0 2
0 2  − 3 x 3 10 x 2 
 6x2  P
x = + + 8x
= + 4 x 0  3 2 0
 2 0
 − 3(2) 3 10(2) 2 
 6(2)
=
2

+ 4(2)  − 0
Area of shaded region =
 3 + + 8(2)  − 0

 2   2 
  =28 - 20
=28 unit2.
=20 unit . 2 = 8 unit 2.

16. Finding the area under a curve
using definite integral
Find the coordinate of P and Q.
Hence, find the area of the shaded region
y Coordinates P and Q.
y=x+4 y=x+4 … …(1)
y = (x - 2 ) 2 …….. ( 2 )
Q
eq (1) = eq (2)
y = (x – 2)2 x+4=(x–2)2
P
x2 - 4x + 4 - x - 4 = 0
x
0
x2 - 5x = 0
x(x – 5) = 0
subtitute x =0 into eq(1)
x = 0 or x = 5
y=0+4=4
subtitute x = 5 into eq (1)
y=5+4=9
coordinate P(0 , 4) and Q(5, 9)

17. Finding the area under a curve
using definite integral
Find the coordinate of P and Q.
Hence, find the area of the shaded region
y
y=x+4
Q
y = (x – 2)2
P
x
0

18. Finding the area under a curve
Area of shaded region
using definite integral
y
y=x+4
Q
y = (x – 2)2
P
x Area under straight line
0 2
Area under curve Area = ∫ ydx
2 5
0
Area = ∫ ydx = ∫ ( x + 4)dx
0 0 5
5  x2 
= ∫ ( x − 2) 2 dx =  + 4 x
2 0
0 5
 ( x − 2) 3   (5) 2 
= =
 2 + 4(5)  − 0

  
 3(1)  0
 (5 − 2) 3 (0 − 2)3  = 65/2 unit2.
=
 3 − 3 −0 
 
=35/3 unit2.

19. Finding the area under a curve
Area of shaded region
using definite integral
y y
y=x+4
Q
y = (x – 2)2
P
x x
0 y 0
Area under curve Area under straight line
2 2
Area = ∫ ydx Area = ∫ ydx
0 5
0
= ∫ ( x + 4)dx
5
= ∫ ( x − 2) 2 dx
0 5
0 5
x  x2 
 ( x − 2) 3  0 =  + 4 x
=  2 0
 3(1)  0
Area of shaded region  (5) 2 
 (5 − 2) 3 (0 − 2)3  =
 2 + 4(5)  − 0

=
 3 − 3 −0   
  = 65/2 - 35/3 .
= 205/6 unit2. = 65/2 unit2.
= /3 unit .
35 2

20. y
3
y2 = 9 - x
2
2x + 3y = 6
x
3 9
Diagram shows sebahagian daripada curve
y2 = 9 – x and straight line 2x + 3y = 6.
Calculate the shaded region

21. y
3 y2 = 9 - x
2
2x + 3y = 6
3 9 x
Area of shaded region = area under curve – area of triangle
3
1
= ∫ xdy − × 3 × 2
0
2
3
= ∫ (9 − y 2 )dy - 6
0
3
 y 3
= 9 y −  - 6
 3 0
33
= 9(3) − − 3 = 15
3