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Integration

Integration. The principles of integration were formulated independently by Isaac Newton and Gottfried Leibniz in the late 17th century. Just as differentiation is associated with rates of change, (gradients of tangents), integration is related to areas under curves.

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Integration

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  1. Integration The principles of integration were formulated independently by Isaac Newton and Gottfried Leibniz in the late 17th century.

  2. Just as differentiation is associated with rates of change, (gradients of tangents), integration is related to areas under curves.

  3. Integrating is the opposite of differentiating, so: differentiate integrate But: differentiate integrate Integrating 6x….......which function do we get back to?

  4. The derivative of any constant is 0 So when we reverse the process, ie integrate, there is no way to know if there was a constant in the original expression or what it was. Constant of Integration …………+C is always added and more information is needed before it can be determined

  5. Integration Notation means “integrate 6x with respect to x” means “integrate f(x) with respect to x” This notation was “invented” in 1675 by the German mathematician Gottfried Wilhelm von Leibniz

  6. Integration Integration is the inverse process (opposite) of differentiation Differentiating Integrating Increase power by 1 Multiply by power Divide by power Decrease power by 1 Add C ?

  7. = Add 1 x6 + c 6 ÷ by new power

  8. Just like differentiation, we must arrange the function as a series of powers of x before we integrate.

  9. Determine the function f (x) We also know f (2) = 11

  10. A curve has a derivative  Given that y passes through (8 , 8) find the y coordinate when x = 1. Since (8 , 8) lies on curve y = 8 when x = 8

  11. Area under a Curve The integral of a function can be used to determine the area between the x-axis and the graph of the function. This is a definite integral, with lower limitaand upper limitb.

  12. F(5) – F(1)

  13. F(2) – F(0)

  14. Very Important Note: y = f(x) + c d a b When calculating integrals: areas above the x-axis are positive areas below the x-axis are negative When calculating the area between a curve and the x-axis: • make a sketch • calculate areas above and below the x-axis separately • ignore the negative signs and add

  15. The Area Between Two Curves To find the area between two curves we evaluate: Top curve Bottom curve a and b are the x-coordinates of the points at which the curves intersect

  16. Calculate the area enclosed by the lines x = 2, x = 4 and the curves y = x2 and y = 4 – x2

  17. The cargo space of a small bulk carrier is 60m long. The shaded part of the diagram represents the uniform cross-section of this space. 9 Find the area of this cross-section and hence find the volume of cargo that this ship can carry. 1

  18. The shape is symmetrical about the y-axis. So we calculate the area of one of the light shaded rectangles and one of the dark shaded wings. The area is then double their sum. The rectangle: let its width be s The wing: extends from x = s to x = t The area of a wing (W ) is given by:

  19. The area of a rectangle is given by: The area of the complete shaded area is given by: The cargo volume is:

  20. 4x + y + 2 = 0 A jeweller has sketched an earring on coordinate axes, with the boundary lines as shown. Calculate its area. y = x2 + 2 y = 2x + 1 –2 –0∙5 1 Step 1: Find the x-coordinates of the intersection points Using y = – 4x – 2 andy = x2 + 2 givesx2 + 2 = – 4x - 2 x2 + 4x + 4 = 0  (x + 2)2 = 0 x = –2 Using y = 2x + 1 andy = x2 + 2 givesx2 + 2 = 2x + 1 x2 – 2x + 1 = 0  (x – 1)2 = 0 x = 1 Straight lines cut x-axis at 2x + 1 = 0 x = –0∙5

  21. Step 2: Split area into two parts. 4x + y + 2 = 0 TOP – BOTTOM y = x2 + 2 y = 2x + 1 –2 –0∙5 1 2 1

  22. 4x + y + 2 = 0 y = x2 + 2 y = 2x + 1 –2 –0∙5 1  Area = 1∙25 sq.units Area = 0∙125 + 1∙125

  23. Questions from Past Papers

  24. The graph of y = g(x) passes through the point (1 , 2) express y in terms of x. When x = 1, y = 2

  25. passes through the point (–1, 2). A curve for which Express y in terms of x. Use the point

  26. 2011 Questions

  27. Find

  28. Find

  29. y = x3 – x2 – 4x + 4 y = 2x + 4 –2 3 The curve and the line intersect at the points (–2, 0), (0, 4) and (3, 10). Calculate the total area between the curves. TOP – BOTTOM

  30. y = x3 – x2 – 4x + 4 y = 2x + 4 –2 3 The curve and the line intersect at the points (–2, 0), (0, 4) and (3, 10). Calculate the total area between the curves. TOP – BOTTOM

  31. More Past Paper Questions

  32. Find the shaded area. Area negative since below x-axis.

  33. The parabola shown in the diagram has equation y = 32 – 2x2. Calculate the shaded area. –2 2 TOP – BOTTOM –3 3 Curves intersect where y = y 32 – 2x2 = 24 32 – 2x2 = 14 8 = 2x2 18 = 2x2 x2 = 4 x2 = 9 x= ± 2 x= ± 3

  34. The parabola shown in the diagram has equation y = 32 – 2x2. Calculate the shaded area.

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