MEI C2 Study Resources Core2 Integration 1-2 by Bob Sheehan

By Bob Sheehan

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Working with other students will help to highlight any misconceptions you may have, as well as making the task more enjoyable. At first glance this may appear to be an easy undertaking, but you will find it takes quite a lot of thought and errors can be made very easily. Make sure you check each coupling, or you could be left with one or two pieces that appear to fit nowhere. 75 ( x  2)2 dx = (a) 0 (b) 8 23 (c) 4 (d) 8 23 (e) I don’t know 3)  1 2 ( x 2  4 x) dx  (a) 3 (c) 7 23 (e) I don’t know (b) -3 (d) 7 23 4) The area enclosed by the curve y = 4x  x2 and the x axis is (b) 0 (a) 10 23 (c) 10 (e) I don’t know 2 3 (d) 32 5) The area enclosed by the curve y = x2  1 and the x-axis is: (a) 1 13 (c) 0 (b) 1 13 (d) 2 3 (e) I don’t know © MEI, 26/08/09 1/3 MEI C2 Integration Section 2 MC test 6) The area in square units enclosed by the curve y = x3  x2 – 2x and the positive x-axis is (a) 8 3 37 12 (b) (c) (e) I don’t know (d) 9 4 5 12 7) The area in square units enclosed by the curve y = x2 + x  2 and the lines y = 0, x = 0 and x = 2 is: (a) 0 (b) 1 56 (d) 3 (c) 23 (e) I don’t know 8) The diagram below shows the curve y  f ( x) .

2. [4] (iii) Use a sketch to explain why the trapezium rule gives an overestimate of the true area. [2] (iv) Calculate the percentage error in the calculation.

75 square units 4. y  x( x  1) The graph cuts the x-axis at the origin and the point (1, 0). y This area is negative as it is below the x-axis. x 1 2 © MEI, 29/06/10 The areas above and below the x-axis must be calculated separately. 2/5 MEI C2 Integration Section 2 Exercise solutions 1 Area below x-axis   x( x  1)dx 0 1   ( x 2  x )dx 0 1   31 x 3  21 x 2  0  31  21   61 2 Area above x-axis   31 x 3  21 x 2  1   83  2    31  21   56 Total area  61  56  1 square unit.

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