Page 102 - DMTH202_BASIC_MATHEMATICS_II

Unit 7: Definite Integral Applications

Step 2: Find out the points of intersection. Notes
If you did not recognize the points of intersection from the graph, solve for them algebraically
or by means of your calculator. To locate them algebraically, set each equation equivalent to
each other.
2
2
2
4 – x = x – 4  – 2x = – 8  x = 4  x = – 2 or x = 2
2
Step 3: Set up and estimate the integral.
Recollect from previous notes, when we were locating the area between the curve and the x-axis,
we had to find out the upper and the lower curve. Then the area was defined to be the next
integral.
b
Area  a  upper curve  lower curve  dx
So the definite integral would be as below.
2 2
Area  2     4  x 2   x 2   4  dx  2    8 2x dx 2 
 
Now, let us assess the integral.

2 2 2 2 2

Area    2    4 x   x   4 dx   2  8 2x dx 

2
 2 3   2 3   2 3 

  8x  x    8   2    2    8   2   2  
 3   3   3 
 2
 16   16  64
 
  16    16   
 3   3  3
If you observe the graph of the two functions cautiously, you should have observed that we
could have used some balance when setting up the integral. The region is symmetric regarding
both the x- and the y-axis. If we had utilized the y-axis symmetry, the consequent integral would
have had bounds of 0 and 2, and we would have had to take 2 times the area to discover the total
area. Here is this integral.
2 2

Area  2 0     4 x 2   x 2   4 dx   2  8 2x dx 2 

2
 2 3   2 3   2 3  

 2 8x  x    8   2    2   8   0    0 
 
 3   2   3   3 
 16   32  64
 2 16   2   
 
 3   3  3
If we had used both symmetries, the consequential integral would still have bounds of 0 and 2,
but the upper function would have been f (x) and the lower function would be y = 0 (the x-axis).
To locate the total area, we would have to obtain this area times 4. Here is this integral.
2 2
Area  4 0     4 x 2  0 dx    4 0   4 x dx 2 
2
 1 3   1 3   1 3  
 4 4x  x   4   4   2    2   4   0    0 


 3  0   3   3 
 8   16  64
 4 8    4   

 3  3  3
2
2
Example: Find the area between the curves f (x) = 0.5sec x and g (x) = – 4sin x over the
interval [–  /3,  /3].

LOVELY PROFESSIONAL UNIVERSITY 97