Calculus - Integration - Areas.
Test Yourself 1 - Solutions.
From the x-axis. | 1.
Comment.We add the slice from the x-axis after drawing the graph and then write down our integration statement. |
2. |
3. (i)
(ii)
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4. | |
5.
So when you have easy regular shapes, don't use calculus but use your standard approach to finding an area. |
6. | |
7. (i)
(ii) There are three separate areas here. The middle area is below the x axis and so the area calculated with the integral has a negative value. That will have to be changed using the absolute value sign.
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8. (i)
(ii) Again there are three separate areas with the first and third under the x axis - and so the area will be calculated as negative. Also here we use basic areas and do not use calculus. Area 1 = 0.5 × 2 × 4 = 4 Area 2 = 0.5 × 4 × 4 = 8 Area 3 = 0.5 × 1 × 2 = 1 ∴ Total area = 13 u 2 |
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From the y-axis. | 9.
(i)
(ii) |
10. |
Between 2 curves - 2 points of intersection. | 11. (i)
(ii) The integral has limits of the two points of intersection - so when 2x = 6x - x2. Solutions are x = 0 and x = 4.
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12. (i)
(ii) (iii) |
13. | 14. (i)
(ii) (iii) |
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Between 2 curves - one point of intersection. | 15. | 16. |
Integrals! | 17. |
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18. (i)
(ii) |
(iii) The area must be calculated in two parts - the region below the axis between x = 1 and x = 2 and then the region above the axis between x = 2 and x - 4.
(iv) The area calculated in (iii) is larger than the value for the integral in (ii) because the area requires adding positive terms together. The answer in (ii) has a positive and a negative term added together and hence the total is smaller. If the total area calculated (59/6) has 2 times (7/6) subtracted from it, the result is the answer for the integral in (ii) (15/2). |
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Interpreting diagrams. | 19. (i) | 20. (i) The integral is asking for the area between the curve and the x axis - with signs taken into account.
Area triangle 1 = 0.5 × 1 × 1 = 0.5 Area triangle 2 (base from x = 1 to 3) Area rectangle (base from x = 3 to 5) OR we could have calculated the area of the trapezium rather than the last two shapes separately. So (ii) The first value of a is obtained when the integral is evaluated on the positive side of the x-axis. From the above, the area of the rectangle - 4 and so we could write and hence a = 3. The second value of a is obtained when the integral is evaluated to include the negative side of the x axis. The integral in (i) = 5.5 so we need an area under the x axis = -1.5. The area of the trapezium from x = -2 to 0 equals 0.5 × (2+1)×1 = 1.5 and the integral is zero. Hence 5.5 - 1.5 = 4 as required. The two values for a are -2 and 3. |