Calculus  Integration  Areas.
Test Yourself 1  Solutions.
From the xaxis.  1.
Comment.We add the slice from the xaxis after drawing the graph and then write down our integration statement. 
2. 
3. (i)
(ii)

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.

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} 

From the yaxis.  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  x^{2}. Solutions are x = 0 and x = 4.

12. (i)
(ii) (iii) 
13.  14. (i)
(ii) (iii) 

Between 2 curves  one point of intersection.  15.  16. 
Integrals!  17. 

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). 

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 xaxis. 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. 