How do you use the limit process to find the area of the region between the graph #y=16-x^2# and the x-axis over the interval [1,3]?

1 Answer
Dec 21, 2016

Answer:

Here is a limit definition of the definite integral. (Others are possible.)

#int_a^b f(x) dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax#.

Explanation:

#int_a^b f(x) dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax#.

Where, for each positive integer #n#, we let #Deltax = (b-a)/n#

And for #i=1,2,3, . . . ,n#, we let #x_i = a+iDeltax#. (These #x_i# are the right endpoints of the subintervals.)

I prefer to do this type of problem one small step at a time.

#int_1^3 (16-x^2) dx#.

Find #Delta x#

For each #n#, we get

#Deltax = (b-a)/n = (3-1)/n = 2/n#

Find #x_i#

And #x_i = a+iDeltax = 1+i2/n = 1+(2i)/n#

Find #f(x_i)#

#f(x_i) = 16-(x_i)^2 = 16-(1+(2i)/n)^2#

# = 16-(1+(4i)/n+(4i^2)/n^2)#

# = 15 -(4i)/n - (4i^2)/n^2#

Find and simplify #sum_(i=1)^n f(x_i)Deltax # in order to evaluate the sums.

#sum_(i=1)^n f(x_i)Deltax = sum_(i=1)^n( 15 -(4i)/n - (4i^2)/n^2) 2/n#

# = sum_(i=1)^n( 30/n -(8i)/n^2 - (8i^2)/n^3)#

# =sum_(i=1)^n ( 30/n) - sum_(i=1)^n((8i)/n^2) - sum_(i=1)^n((8i^2)/n^3)#

# =30 /nsum_(i=1)^n ( 1)-8/n^2sum_(i=1)^n(i)-8/n^3sum_(i=1)^n(i^2) #

Evaluate the sums

# = 30/n(n) -8/n^2((n(n+1))/2) - 8/n^3((n(n+1)(2n+1))/6)#

(We used summation formulas for the sums in the previous step.)

Rewrite before finding the limit

#sum_(i=1)^n f(x_i)Deltax = 30/n(n) - 8/n^2((n(n+1))/2) - 8/n^3((n(n+1)(2n+1))/6)#

# = 30 - 4((n(n+1))/n^2) - 4/3((n(n+1)(2n+1))/n^3)#

Now we need to evaluate the limit as #nrarroo#.

#lim_(nrarroo) ((n(n+1))/n^2) = 1#

#lim_(nrarroo) ((n(n+1)(2n+1))/n^3) = 2#

To finish the calculation, we have

#int_0^1 x^2 dx = lim_(nrarroo) (30 - 4((n(n+1))/n^2) - 4/3((n(n+1)(2n+1))/n^3)#

# = 30 - 4(1) - 4/3(2)#

# = 90/3 - 12/3 - 8/3 = 70/3#.