Definite Integral and Area

Aims: By the end of this chapter, you will be able to use definite integral to find
(i) area underneath a curve, and
(ii) area bounded by two curves.

  1. From the previous note on the introduction to the area beneath a curve, we learned that area underneath a curve can be calculated using the definite integral. In this chapter, we look at some examples on how definite integral can be used to find area.
  2. Example 1: Find the shaded area in the diagram shown here (right).
    The curve in this diagram is y = 5-x2.
    The shaded area = ∫ab y(x) dx.
    The shaded area = ∫-12 [5-x2] dx.
      shaded area = [5x - x3/3]-12
    shaded area = [5(2) - (2)3/3] - [5(-1) - (-1)3/3]
    shaded area = [10-8/3] - [-5 +1/3]
    shaded area = 12
  3. Example 2: Find the shaded area in the diagram shown here (right).
    The curve in this diagram is y = 4x3.
    The shaded area = ∫ab y(x) dx.  
    The shaded area = ∫13 [4x3] dx.
      shaded area = [x4]13
    shaded area = [34] - [14]
    shaded area = 81 - 1
    shaded area = 80
  4. Example 3: (a) Find the shaded area in the diagram shown below and report your answer accurate to three significant figures.
    The curve in this diagram (right) is y = 3 cos (x) and x is measured in radian.
    The shaded area in this diagram is actually separated into two parts; one above the x-axis and one below the x-axis. The curve y=3 cos x crosses the x-axis at π/2. If we were to find the area of the part below the x-axis then we need to calculate ∫π/22 3cos(x) dx.
    The shaded area of this part below the x-axis = [ 3 sin (x) ]π/22
      The shaded area = 3 sin (2) - 3 sin (π/2)
    The shaded area ≈ -0.2721077195. We now have a negative area. We know that area cannot be negative. The negative sign basically means that the area is located below the x-axis. Since we are not interested in location but the actual magnitude of the area. We will have to take the absolute value of this area; | -0.2721077195 | ≈ 0.2721077195.
    Total shaded area = ∫0π/2 3cos(x) dx +| ∫π/22 3cos(x) dx |
    Total shaded area ≈ [3 sin x]0π/2 +   | -0.2721077195 |
    Total shaded area ≈ 3 sin (π/2) - 3 sin (0) + 0.2721077195
    Total shaded area ≈ 3(1) - 0 + 0.2721077195
    Total shaded area ≈ 3.27 (3 s.f.)
  5. If the shaded area of a curve y(x) is separated into two parts at x = c such that one part is above the x-axis and the other is below that x-axis then the total shaded area is
        ∫ac y(x) dx + | cb y(x) dx |
    or
        ∫ac y(x) dx + ( -cb y(x) dx )
    since the value of the second part is negative then a negative sign in front would transform the whole thing into a positive value.

    In the example 3 above, the value of c is where the curve y(x) crosses the x-axis. The value c is π/2 in this particular example.

    GDC

    Besides using the function fnInt( function to find definite integral. We can also use graphical method. In this case, enter our equation into [Y= ]. We then plot the graph using [GRAPH]. To calculate a definite integral as in example 1 above, we press [2nd][TRACE] for CALC and then press 7: ∫ f(x) dx.
    The calculator will prompt us for the lower limit. Enter the lower limit directly. In this case, -1.
    Enter the upper limit next, which is 2 in this particular example.
    Our graph will be shaded in the appropriate region and the answer is displayed at the bottom of the screen.,12 in this case.

  6. Try this question. Find the area of the region bounded by y = 1/x, x = e1/4, x = e and y = 0.2.
    You may want to start this problem by sketching out this region.
     

     

     

     

     

    Key: 0.463 (3.s.f.)

  7. Try this one. Two functions f(x) and g(x) are given as f(x) = (x-2)4 and g(x) = 4 -(x-2)2.
    (a) Sketch f(x) and g(x) on the same axes.
    (b) Function f(x) intersects g(x) at two points. Find these intersection points and report answers accurate to three significant figures.
    (c) Use your reseults from (b) above and find the area of the region bounded by f(x) and g(x).
     

     

     

     

     

     

     

     

    Key: (b) (0.750,2.44) & (3.25,2.44); (c) 7.48 (3 s.f.)

  8. Using the results from above, write the appropriate expressions for calculating
    (a) the area of the little blue region
     
    (b) the area of the larger green region
     
    (c) the area of the total shaded region
  9. Applications to Kinematics

  10. displacement (s)→ [ds/dt] → velocity (v) → [dv/dt] → acceleration (a)

    displacement (s ) ← [ v dt] velocity (v) [ a dt] acceleration (a)

    Displacement is the distance from the origin and has sign. Let say you stand below a tree and call this spot your origin. If you allow steps to the right of the tree to be positive steps then negative steps are steps to the left of the tree.
    "5 steps to the right of the origin" means displacement (s) = 5.
    "3 steps to the left of the origin" means s = -3.
    "5 steps to the right of the origin and then 3 steps to the left" means s = 2. However, in this case, you have travelled a total of 8 steps. Thus, the total distance travalled (usually represented by x) = 8.
    "7 steps to the left of the origin and then 4 steps to the right" means s = - 3 but x = 11.
  11. If the graph in example 3 above represents the velocity (m/s) of a particular object where the x-axis represents time t in second. Then the total distance travelled by the object in the first 2 seconds is the total shaded area. That is, we need to take the magnitude or the absolute value of the negative area. In this case the answer is 3.27m (3.s.f) as calculated above.
    The displacement of the object in the first 2 seconds is easier and calculated directly from
    02 3cos(t) dt which is 2.73 m to 3 s.f..
  12. Example 4. The velocity of an object is v m/s after t seconds, where v = 3 - 5 sin (t/2).
    (a) Find the displacement of the object over the first 6 seconds.
    (b) Find the total distance travelled by the object in the first 6 seconds.
    Solutions.
    It is a good idea to have a sketch of the function v. Here is a screen capture from the Ti calculator.
    (a) Displacement over the first 6 seconds =    6 [3 - 5sin(t/2)] dt.
    ≈ -1.90 (3 s.f.)
    (b) The roots in the first 6 seconds are 1.287 and 4.996 to 3 d.p.
    Total distance = 01.287 [3 - 5sin(t/2)] dt +1.2874.996 [3 - 5sin(t/2)] dt + 4.9966 [3 - 5sin(t/2)] dt.
    Total distance = 1.861 + 4.872457 + 1.1115257
    Total distance ≈ 7.84 (3.s.f.)