Trigonometric functions - Integration - Applications finding areas.
Test Yourself 1 - Solutions.
Solve the following problems as indicated:
Area - 1 curve |
1.
The curve is symmetrical about y = π/2. So the area between 0 and π/2 can be determined and then doubled. |
2. (i)
(ii) When y = 1:
∴ There are two shapes: 1. a rectangle of height 1 and length π/4. 2. the area under the curve between π/4 and π/3. For the 2nd area: |
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3. | 4. (iii) |
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4. f(x) = 2cos x + 1,
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5. (i)
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(ii) | |||||||||
6. |
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7. (i)
(ii)
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(iii) (iv) |
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8. | ||||||||||
Area - 2 curves with no point of intersection. |
9. (i)
(ii) |
10. (i) (ii) |
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Area - 2 curves with one point of intersection |
11. | (ii) | ||||||||
12.
(i)
(ii) |
(iii) There are two parts to integrate -
∴ Area = 1 + ln 2 |
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13. (i) | (ii) The two curves are symmetrical about x = π/2. So areas are doubled to obtain total areas from |
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14. (i) | (ii) | |||||||||
Area - 2 curves with two points of intersection |
15.
(i) |
(ii) | ||||||||
16. (i)
x = 0, y = 2sin x = 2 sin 0 = 0
x = 0.896, y = 2sin x = 2 sin 0.896 = 1.562
Hence the curves intersect at x = 0 and |
(ii) |