Please help with this wrong answer? (Circled in red) EXAMPLE 2For the region under f(x) = 5x2 on [0, 21, show that the sum of the areas ofthe upper approximating rectangle approachesthat is40lim Rnn 00SOLUTIONRn is the sum of the areas of the n rectangles in the figure. Each rectangle haswidthand the heights are the values of the function f(x) = 5x2 at the points46...,"; that is, the heights aren'2nThus,2.2.5Rn =++... +n410· (12 + 22 + 32 ++ n?)n240(12 + 22 + 32 + ... + n?).n°„3Here 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 get40Rn =n340(n +1)(2n + 1)6n2n(n + 1)(2n + 1)Thus we have40n2 + 60n + 20lim Rn = limn 003n?20 2+= limn- 00n2n +)12n + 1in320 +0+0(1 + +)(2 + +)= lim32040•1·2 =

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Description: Please help with this wrong answer? (Circled in red) EXAMPLE 2For the region under f(x) = 5x2 on [0, 21, show that the sum of the areas ofthe upper approximating rectangle approachesthat is40lim Rnn 00SOLUTIONRn is the sum of the areas of the n rect

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Answered: 40n“ + 60n + 20 lim lim Rn 3n2

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter3: Functions And Graphs

Section3.2: Graphs Of Equations

Problem 78E

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Please help with this wrong answer? (Circled in red)

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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