Definite Integrals 233
Example
Evaluate
∑n
i= 1
i(i+1)
n
.
Rewrite:
∑n
i= 1
i(i+1)
n
as
1
n
∑n
i= 1
(i^2 +i)=
1
n
( n
∑
i= 1
i^2 +
∑n
i= 1
i
)
=
1
n
(
n(n+1)(2n+1)
6
+
n(n+1)
2
)
=
1
n
[
n(n+1)(2n+1)+ 3 n(n+1)
6
]
=
(n+1)(2n+1)+3(n+1)
6
=
(n+1)[(2n+1)+ 3 ]
6
=
(n+1)(2n+4)
6
=
(n+1)(n+2)
3
.
TIP • Remember: In exponential growth/decay problems, the formulas are dy
dx
=kyand
y=y 0 ekt.
Definition of a Riemann Sum
Let f be defined on [a,b] andxibe points on [a,b] such thatx 0 =a,xn=b, anda <
x 1 <x 2 <x 3 ···<xn− 1 <b. The pointsa, x 1 , x 2 , x 3 ,...xn+ 1 , andbform a partition of
f denoted asΔon [a,b]. LetΔxibe the length of theith interval [xi− 1 ,xi] andcibe any
point in theith interval. Then the Riemann sum off for the partition is
∑n
i= 1
f(ci)Δxi.
Example 1
Letf be a continuous function defined on [0, 12] as shown below.
x 0 2 4 6 8 10 12
f(x) 3 7 19 39 67 103 147
Find the Riemann sum for f(x) over [0, 12] with 3 subdivisions of equal length and the
midpoints of the intervals asci.
Length of an intervalΔxi=
12 − 0
3
=4. (See Figure 11.1-1.)