5 Steps to a 5 AP Calculus AB 2019 - William Ma

(Marvins-Underground-K-12) #1
MA 3972-MA-Book April 11, 2018 15:57

Def inite Integrals 249

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

.


(Note: This question has not appeared in an AP Calculus AB exam in recent years.)

TIP • Remember: In exponential growth/decay problems, the formulas are dy
dx
=kyand
y=y 0 ekt.


Definition of a Riemann Sum
Letfbe 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 12.1-1.)
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