Definite Integrals

Example 1. f(x) = x2, a = 0, b arbitrary
1. Divide into n intervals
Length b/n = base of rectangle
2. Heights:
? ?2 b b • 1st: x = , height =
n n
? ?2b 2b 2 • 2nd: x = , height =
n n
Sum of areas of rectangles:
? ?? ?2 ? ?? ?2 ? ?? ?2 ? ?? ?2 b b b 2b b 3b b nb b3
+ + + + = (12 + 22 + 32 + + n2)
n n n n n n
· · ·
n n n3
· · ·
1
Lecture 18 18.01 Fall 2006
a=0 b
Figure 2: Area under f(x) = x2 above [0, b].
We will now estimate the sum using some 3-dimensional geometry.
Consider the staircase pyramid as pictured in Figure 3.
n = 4
n
Figure 3: Staircase pyramid: left(top view) and right (side view)
1st level: n × n bottom, represents volume n2.
2nd level: (n ? 1) × (n ? 1), represents volumne (n ? 1)2), etc.
Hence, the total volume of the staircase pyramid is n2 + (n ? 1)2 + · · · + 1.
Next, the volume of the pyramid is greater than the volume of the inner prism:
1 1 1
12 + 22 + · · · + n2 > (base)(height) = n2 n = n3 3 3 · 3
and less than the volume of the outer prism:
1 1
12 + 22 + · · · + n2 < (n + 1)2(n + 1) = (n + 1)3 3 3
2
? ?
? ?
Lecture 18 18.01 Fall 2006
In all,
1 1n3 12 + 22 + + n2 1 (n + 1)3
= 3 <
· · ·
< 3 n3 n3 3 n3
Therefore,
b3 1
lim (12 + 22 + 32 + + n2) = b3,
n?? n3
· · · 3
b3
and the area under x2 from 0 to b is .
3
Example 2. f(x) = x; area under x above [0, b]. Reasoning similar to Example 1, but easier, gives
a sum of areas:
b2 1
n2 (1 + 2 + 3 + · · · + n) ? 2b2 (as n ? ?)