| From the x-axis. |
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2. |
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3.  |
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4. |
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5.  |
6.  |
| From the y-axis. |
7. (i) 
(ii) We first write the integral statement for the area:
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We now take the negative part of this y equation as we want the part of the parabola below x = 2:
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8. |
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9. |
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| Between 2 curves. |
10. 
Need the x-values for the point of intersection:
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11. (i) Verify one of the points of intersection has coordinates (2, 1).
Equating the y expressions:
x2 - 2x + 1 = 3 - x
x2 - x - 2 = 0
(x - 2)(x + 1) = 0
x = 2 or x = -1
x = 2 y = 1 So POI is (2, 1).
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(ii) Hence find the area enclosed.
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12. (i) 
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(ii)  |
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13.  |
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14. 
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| Between 2 curves - one point of intersection. |
15. |
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16. |
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| Integrals! |
17. (i)  |
(ii)
(iii) Curve is not differentiable at x = 2 and at x = 4.
Note at x = 6, the gradient of CD flows into the gradient of the quadrant. |
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18. |
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| Interpreting diagrams. |
19.  |
20. |