For the region under f(x) 5x on [0, 2], show that the sum of the areas of the upper approximating rectangle approaches XAMPLE 2 that is 3 40 lim Ra n- 00 R, is the sum of the areas of the n rectangles in the figure. Each rectangle has width 2 and the heights are the values of the OLUTION unction f(x) = 5x2 at the points 2, 4, 5 ,2n, that is, the heights are (2). Thus, R., = () +2- () + 2 - () + 10. • (12 + 22 + 32 + ... + n?) (12 + 2? ² + ... + n²). ere we need the formula for the sum of the squares of the first n positive integers: 12 + 22 + 32 +.. + n? = nln + 1)(2n + 1) erhaps you have seen this formula before. Putting this formula into our expression for Rp, we get n(n + 1)(2n + 1) hus we have lim R. lim %3D %3D lim n- 00 = lim n- 00 (1 + )(2 + ÷) •1:2 =
For the region under f(x) 5x on [0, 2], show that the sum of the areas of the upper approximating rectangle approaches XAMPLE 2 that is 3 40 lim Ra n- 00 R, is the sum of the areas of the n rectangles in the figure. Each rectangle has width 2 and the heights are the values of the OLUTION unction f(x) = 5x2 at the points 2, 4, 5 ,2n, that is, the heights are (2). Thus, R., = () +2- () + 2 - () + 10. • (12 + 22 + 32 + ... + n?) (12 + 2? ² + ... + n²). ere we need the formula for the sum of the squares of the first n positive integers: 12 + 22 + 32 +.. + n? = nln + 1)(2n + 1) erhaps you have seen this formula before. Putting this formula into our expression for Rp, we get n(n + 1)(2n + 1) hus we have lim R. lim %3D %3D lim n- 00 = lim n- 00 (1 + )(2 + ÷) •1:2 =
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Example 2:** For the region under \( f(x) = 5x^2 \) on \([0, 2]\), show that the sum of the areas of the upper approximating rectangles approaches \(\frac{40}{3}\), that is
\[
\lim_{{n \to \infty}} R_n = \frac{40}{3}
\]
**Solution:** \( R_n \) is the sum of the areas of the rectangles in the figure. Each rectangle has width \(\frac{2}{n}\) and the heights are the values of the function \( f(x) = 5x^2 \) at the points \(\frac{2}{n}, \frac{4}{n}, \frac{6}{n}, \ldots, \frac{2n}{n}\), that is, the heights are \(\left( \frac{2}{n} \right)^2, 5 \left( \frac{4}{n} \right)^2, 5 \left( \frac{6}{n} \right)^2, \ldots, 5 \left( \frac{2n}{n} \right)^2 \).
Thus,
\[
R_n = \frac{2}{n} \cdot 5 \left( \frac{2}{n} \right)^2 + \frac{2}{n} \cdot 5 \left( \frac{4}{n} \right)^2 + \ldots + \frac{2}{n} \cdot 5 \left( \frac{2n}{n} \right)^2
\]
\[
= \frac{10}{n^3} \left(1^2 + 2^2 + 3^2 + \ldots + n^2 \right)
\]
\[
= \frac{10}{n^3} \left(1^2 + 2^2 + 3^2 + \ldots + n^2 \right).
\]
Here we need the formula for the sum of the squares of the first \( n \) positive integers:
\[
1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6}.
\]
Perhaps you have seen this](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5abad2ef-7d64-4544-b6f7-dd3dfbff74f8%2Fabb944b8-d822-4720-aece-eac5e1358301%2Fh3e7v6_processed.png&w=3840&q=75)
Transcribed Image Text:**Example 2:** For the region under \( f(x) = 5x^2 \) on \([0, 2]\), show that the sum of the areas of the upper approximating rectangles approaches \(\frac{40}{3}\), that is
\[
\lim_{{n \to \infty}} R_n = \frac{40}{3}
\]
**Solution:** \( R_n \) is the sum of the areas of the rectangles in the figure. Each rectangle has width \(\frac{2}{n}\) and the heights are the values of the function \( f(x) = 5x^2 \) at the points \(\frac{2}{n}, \frac{4}{n}, \frac{6}{n}, \ldots, \frac{2n}{n}\), that is, the heights are \(\left( \frac{2}{n} \right)^2, 5 \left( \frac{4}{n} \right)^2, 5 \left( \frac{6}{n} \right)^2, \ldots, 5 \left( \frac{2n}{n} \right)^2 \).
Thus,
\[
R_n = \frac{2}{n} \cdot 5 \left( \frac{2}{n} \right)^2 + \frac{2}{n} \cdot 5 \left( \frac{4}{n} \right)^2 + \ldots + \frac{2}{n} \cdot 5 \left( \frac{2n}{n} \right)^2
\]
\[
= \frac{10}{n^3} \left(1^2 + 2^2 + 3^2 + \ldots + n^2 \right)
\]
\[
= \frac{10}{n^3} \left(1^2 + 2^2 + 3^2 + \ldots + n^2 \right).
\]
Here we need the formula for the sum of the squares of the first \( n \) positive integers:
\[
1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6}.
\]
Perhaps you have seen this
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