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 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F998b6cca-58bb-4391-bb13-94678d180001%2Fd1b97f7e-5af8-4d8d-9842-a133b63e24cc%2F6oc0o68_processed.png&w=3840&q=75)
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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