Intermediate Answers to Exercise 10.1-2

The base is length a, so if you divide it into n equal slices, each slice will be a/n wide.

The figure on the right will help you visualize how we are slicing the area under the parabola. Here we are cutting it into four slices -- that is n = 4 (note that the leftmost slice has height of zero). The first slice has its left-hand edge at x = 0. The second at x = a/n. The third at x = 2a/n. That pattern continues. So the left-hand edge of the kth slice is at x = (k-1)a/n.

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Recall that f(x) = x². And the result from step 2 is that the left-hand edge of the kth rectangle is at x = (k-1)a/n. So just substitute the x into the f(x) with that expression to get the height of the kth rectangle, h_k:

   h_k  =  f((k-1)a/n)  =  (k-1)²a²/n²

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The width of every slice is a/n. The height of the kth slice is (k-1)²a²/n². So the area, A_k, of the kth slice is width times height, which is:

   A_k  =  (k-1)²a³/n³

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Just put the expression you got in step 4 into a summation for k equal 1 to n.

          n         n
   A_total  =   å A_k  =   å (k-1)²a³/n³
         k=1       k=1

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The factor that is constant with respect to k is a³/n³. When you factor it out you get

               n
   A_total  =  (a³/n³) å (k-1)²
              k=1

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The summation from step 6 is just the sum of the squares of the integers from zero to n-1. We have a formula for that. When you apply it you get

               n(n-1)(2n-1)
   A_total  =  (a³/n³)             
                    6

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When you multiply out the polynomial you get

               2n³ - 3n² + n
   A_total  =  (a³/n³)              
                    6

Now multiply through by the a³/n³:

         a³      a³      a³
   A_total  =      -       +     
          3      2n     6n²

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The terms with n in them all have the n in the denominator. So as n gets very big, these terms become closer and closer to zero, hence they become insignificant. All that remains is the term with no n in it. So in the limit you have

   A_total  =  a³/3

which is the actual area of the shaded region that we started with.

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How to find the sum of the squares of the integers from zero to n-1: First observe what the difference of two consecutive cubes is:

   n³ - (n - 1)³   =  n³  -  (n³ - 3n² + 3n - 1)  =  3n² - 3n + 1

Notice that we had to apply the binomial formula to unravel that.

Now observe that

   n³  =  (1³ - 0³) + (2³ - 1³) + (3³ - 2³) +  ...  

            ... + ((n-1)³ - (n-2)³) + (n³ - (n-1)³)

which works simply because everything cancels except the n³. But isn't that just the same as saying

         n
   n³  =  å k³ - (k-1)³
        k=1

Now, turning the equation around and substituting the difference expression we got using the binomial formula, we have:

    n
    å 3k² - 3k + 1  =  n³
   k=1

Try this with a few small values of n to convince yourself that it works.

The expression in the summation is itself a sum of three things. So we separate it into three summations:

    n         n         n
    å 3k²  -  å 3k  +  å 1  =  n³
   k=1       k=1       k=1

Now factor the constant multipliers from each summation:

     n          n        n
   3 å k²  -  3 å k  +  å 1  =  n³
    k=1        k=1      k=1

Starting with the easiest summation, the sum from 1 to n of 1 added to itself is pretty easy. That's equal to n.

The sum the integers from 1 to n we can do the same we we did in the main text for adding integers from zero to n-1 (by writing the sum forward and backward):

   S  =   1   +   2   +   3   +  ...  +  n-2  +  n-1  +   n
   S  =   n   +  n-1  +  n-2  +  ...  +   3   +   3   +   1
                                                            
  2S  =  n+1  +  n+1  +  n+1  +  ...  +  n+1  +  n+1  +  n+1

Clearly twice the sum is n+1 added to itself n times. So

    n       n(n+1)
    å k  =        
   k=1        2

So substituting for the two sums that we know, our equation becomes:

     n         n(n+1)   
   3 å k²  -  3         +  n  =  n³
    k=1          2

Now move all the stuff that we have resolved so far over to the right and divide by 3

    n        n³     n(n+1)     n
    å k²  =     +          -   
   k=1        3       2       3

This gives the sum of the square integers from 1 to n. But we wanted the sum from zero to n-1. So we have to add zero squared and subtract n². Of course zero squared is zero, so we ignore that. Here is the result:

   n-1       n³     n(n+1)     n
    å k²  =     +          -     -  n²
   k=0        3       2       3

Now multiply out the n(n+1), and put everything over a common denominator:

   n-1     2n³ + 3n² + 3n - 2n - 6n²     2n³ - 3n² + n     n(2n² - 3n + 1)
    å k²  =                           =               =                 
   k=0                  6                      6                  6

All that remains for you to do to finish this is to factor the quadratic on the right.

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