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Unit 7: Definite Integral Applications




          is there a symmetry in the graph of the curve between 0 and 2 and 0 and – 2, but the area under  Notes
          the curve from 0 to 2 is identical as the area under the curve from 0 to – 2. Thus, the complete area
          among this curve and the x-axis is twice the area computation between x = 0 and x = 2.  Observe
          the lack of negative numbers in this solution.

                                           2           3
                                         3
                                                                  
                   2                   x         2         4     12  4  16
              A   2  0       2   dx   2 2x       2  2 2          2 4     2     
                     2 (1/2)x 
                                                  
                                                          0 
                                    
                                       6  0      6         3     3  3    3
          In performing area problems gaze for symmetry that will make the problem simpler and cut
          down on the quantity of numbers you have to influence.
                 Example: An investment in a definite mining process is returning an total per month
          that goes with the following formula
                  2
           A   0.25x   14x  300
          where x is the month and A is the amount for that month.  This formula is applicable for 40 months.

          What is the overall return for the first 12 months?
          Solution:
          The curve begins out at $300 for the zero  month and drops off with time. In this problem the
                                           th
          height of a rectangle beneath the curve for every value of x displays the money obtained that
          month and the sum of these rectangles is the total  quantity for  the period  summed. For 12
          months this amount is the area under the curve from the zero  to the  eleven  month, or the
                                                                          th
                                                              th
          integral from 0 to 11.
                                        11
                                              2
                                    T   0   0.25x   14x   300  dx
                                                         11
                                          x 3   x 3    
                                    T    0.25    14    300x 
                                          3     2       0 
                                          11 3  11 2       
                                                          
                                    T    0.25   14    300 11  
                                           3     2         
                                    T   $2564
          Self Assessment

          Fill in the blanks:

          1.   A straight line, y = const, above a distance in x is a ................................... .

          2.   The area of a rectangle is the height multiplied by the ................................... .
          3.   The area of a ...................................is (1/2)(sum of the opposite faces) (height).
          4.   The rectangles can be created in a numerous methods, inside the curve, outside the curve
               or by using a ................................... .
          5.   It actually doesn’t make any variation how the rectangles are created since we are going to
               take the limit by having their width to ................................... .
          6.   By means of a ................................... approach, and the knowledge that this outlines, or
               integral, over a particular range in x is the area under the curve, A is the limit of the sum
               as Dx leads to zero.




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