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
( )