40n“ + 60n + 20 lim lim Rn 3n2

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 94E
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EXAMPLE 2
For the region under f(x) = 5x2 on [0, 21, show that the sum of the areas of
the upper approximating rectangle approaches
that is
40
lim Rn
n 00
SOLUTION
Rn is the sum of the areas of the n rectangles in the figure. Each rectangle has
width
and the heights are the values of the function f(x) = 5x2 at the points
4
6
...,"; that is, the heights are
n'
2n
Thus,
2.
2.5
Rn =
+
+... +
n
4
10
· (12 + 22 + 32 +
+ n?)
n2
40
(12 + 22 + 32 + ... + n?).
n°
„3
Here we need the formula for the sum of the squares of the first n positive integers:
12 + 22 + 32 +... + n2 - n(n + 1)(2n + 1)
Perhaps you have seen this formula before. Putting this formula into our expression for Ro.
we get
40
Rn =
n3
40(n +1)(2n + 1)
6n2
n(n + 1)(2n + 1)
Thus we have
40n2 + 60n + 20
lim Rn = lim
n 00
3n?
20 2+
= lim
n- 00
n2
n +)1
2n + 1
in
3
20 +0+0
(1 + +)(2 + +)
= lim
3
20
40
•1·2 =
Transcribed Image Text:EXAMPLE 2 For the region under f(x) = 5x2 on [0, 21, show that the sum of the areas of the upper approximating rectangle approaches that is 40 lim Rn n 00 SOLUTION Rn is the sum of the areas of the n rectangles in the figure. Each rectangle has width and the heights are the values of the function f(x) = 5x2 at the points 4 6 ...,"; that is, the heights are n' 2n Thus, 2. 2.5 Rn = + +... + n 4 10 · (12 + 22 + 32 + + n?) n2 40 (12 + 22 + 32 + ... + n?). n° „3 Here we need the formula for the sum of the squares of the first n positive integers: 12 + 22 + 32 +... + n2 - n(n + 1)(2n + 1) Perhaps you have seen this formula before. Putting this formula into our expression for Ro. we get 40 Rn = n3 40(n +1)(2n + 1) 6n2 n(n + 1)(2n + 1) Thus we have 40n2 + 60n + 20 lim Rn = lim n 00 3n? 20 2+ = lim n- 00 n2 n +)1 2n + 1 in 3 20 +0+0 (1 + +)(2 + +) = lim 3 20 40 •1·2 =
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