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Area under a graph Car travelling at 70 mph Area = This is the distance travelled, 140 miles 2 70 = 140 v mph t hours 0 2 70

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Car accelerating steadily from 0 to 45 mph in 10 seconds Area = Distance travelled = 110 yards = 110 v mph t seconds 0 10 45 45 mph = = 22 yards per second 22

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Car travelling between 2 sets of traffic lights 0 Area of A = t (s) v (ms -1 ) 0 0 2 5 4 8 8 8 10 5 12 0 6 9 v ms -1 t seconds 12 624810 A A CC B B = 70 Area of C == 17 Area of B = = 13 Total Area = 5 Distance travelled = 70 metres

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y x 0 a b Integration As x 0 Integration is the inverse of differentiation. lim x 0 A = As A = y x A = xx y y x AA The gradient of the graph of A against x.

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Rules of Integration = x y = 5x = 5 y = c = 0 Function Derivative Reverse to give integral = 5x y = = x 2 y = = x 3 y = + c

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13 Example Find the area under y = 2x 3 - 9x 2 + 12x + 6 between x = 1 and x = 3 y x 0 y = 2x 3 - 9x 2 + 12x + 6 A = + 6x - 3x 3 + 6x 2 + 6 3 -3 3 3 + 6 3 2 + 6 1 -3 1 3 + 6 1 2 = 31.5- 9.5 = 22 square units

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Distance = The distance travelled is 140 miles = 140 Car travelling at 70 mph v mph t hours 0 2 70 Velocity-time graphs v = 70

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Car accelerating steadily from 0 to 22 yd/s (45 mph) in 10s v yd/s t seconds 0 10 22 gradient = v = 2.2t intercept = 0 = 2.2 Distance = The distance travelled is 110 yards = 110

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Car travelling between 2 sets of traffic lights t024810126 v0588509 v ms -1 t seconds 0 12 Distance travelled = 72 metres v = 0.25t(12 - t) Distance = = 72

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5.1 Accumulated Changes Example 1: An objects travels with a velocity of 15 mph. What is the distance traveled after 4 hours t v 15 1234 Distance = area.

5.1 Accumulated Changes Example 1: An objects travels with a velocity of 15 mph. What is the distance traveled after 4 hours t v 15 1234 Distance = area.

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