C. DEFINITION OF DEFINITE INTEGRAL AS THE LIMIT OF
A RIEMANN SUM
Most applications of integration are based on the FTC. This theorem provides
the tool for evaluating an infinite sum by means of a definite integral. Suppose
that a function f(x) is continuous on the closed interval [a,b]. Divide the interval
into n subintervals of lengths Δxk (it is not necessary that the widths be of equal
length, but the formulation is generally simpler if they are). Choose numbers,
one in each subinterval, as follows: x 1 in the first, x 2 in the second, . . . , xk in the
k th, . . . , xn in the nth. Then , where is called
a Riemann Sum.
If we choose equal width subintervals on the interval [a,b], then Δxk is the
same for all values of k, and we use for each width. Given that the set
{x 0 , x 1 , x 2 , . . . , xn} partitions the interval [a,b] into n equal subintervals where
x 0 = a, xn = b and each of x 1 , x 2 , . . . , xn are the x value for the right edge of each
subinterval, then xk = a + k · Δx for each k. We can now write the limit as a
definite integral: .
Example 22 __
Write as a definite integral.
SOLUTION #1: Recall that this limit is of the form · Δx. If we
assume that then a = 2 and , so b = 7; xk replaced x in f(x)
so f(x) = x^2 . This gives us the definite integral .
SOLUTION #2: We could also assume that , then a = 0 and ,
so b = 5 ; xk replaced x in f(x) so f(x) = (2 + x)^2 . This gives us the definite integral
. Indeed, both definite integrals yield the same value: .