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Integration – Overall Objectives. Integration as the inverse of differentiation Definite and indefinite integrals Area under the curve. Area Below Axis. 4. =. 1. 4. 1. y = -x 2 - 3. What is the area between the curve and the x-axis between x=1 and x=4?.
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Integration – Overall Objectives • Integration as the inverse of differentiation • Definite and indefinite integrals • Area under the curve
Area Below Axis 4 = 1 4 1 y = -x2 - 3 What is the area between the curve and the x-axis between x=1 and x=4? = (-64/3 -12) – (-1/3 -3) = -100/3 – -4/3 = -96/3 = -32
The definite Integral • Area under a curve between x=a and x = b is given by the definite integral • Areas below the x axis are negative
Area between line and curve - 1 1 4 y = x2 – 3x + 5 y = 2x + 1 What is the area under the curve between x=1 and x=4?
Area between line and curve - 2 1 4 y = x2 – 3x + 5 y = 2x + 1 What is the area under the curve between x=1 and x=4? 1) Area Under curve 4 = 1 = (64/3 – 24 + 20) – (1/3 -3/2 + 5) = (52/3 – 23/6) = 27/2
Area between line and curve - 2 x = 1 y = 2 + 1 y = 3 (4,9) (1,3) x = 4 y = 8 + 1 y = 9 1 4 Area triangle = 0.5 x (4-1) x (9-3) = 0.5 x 3 x 6 = 9 Area rectangle = (4-1) x 3 = 3 x 3 = 9 y = x2 – 3x + 5 y = 2x +1 2) Area Under Line TOTAL AREA = 9 + 9 = 18
Area between line and curve - 3 1 4 y = x2 – 3x + 5 y = 2x + 1 3) Find difference What is the area under the curve between x=1 and x=4? Area under curve = 27/2 Under line = 18 Area between = line – curve = 18 – 27/2 = 9/2
AQA January 2006 Core 1
4 3 (2, 4) Coordinate C: when x=2; y = 3x2 – x3 = 3x22 – 23 = 12 – 8 = 4 Area of rectangle = length x width = (2 - -1) x 4 = 3 x 4 = 12
2 = -1 B ii) = (23 – ¼ 24) – ((-1)3 – ¼ (-1)4) = (8 - 4) – (-1 – ¼ ) = 5 ¼
B ii) Area of rectangle ABCD = 12 Area under curve = 5 ¼ Shaded area is 12 – 5 ¼ = 6 ¾
AQA January 2006 Core 1
Integration • Exercise C on page 148 • Q 1, 6, 7 Papst Papers + Mark Schemes http://www.aqa.org.uk/qual/gceasa/mathematics_assess.html