Chapter 4: Q. 44 (page 340)

In Exercises 39–44, write out the sigma notation for the Riemann sum described in such a way that the only letter which appears in the general term of the sum is k. Don’t calculate the value of the sum; just write it down in sigma notation.
$f (x) = \sqrt{1 - x^{2}}, [a, b] = [- 1, 1], m i d p o i n t s u m, n = 20$

Short Answer

Expert verified

The midpoint sum is $\frac{1}{10} \sum_{k = 1}^{20} \sqrt{1 - {(\frac{4 k - 13}{20})}^{2}}$

Step by step solution

Step 1. Given Information

We have $f (x) = \sqrt{1 - x^{2}}, [a, b] = [- 1, 1], m i d p o i n t s u m, n = 20$

Step 2. Finding ∆x and xk-1

$∆ x = \frac{b - a}{n} = \frac{1 + 1}{n} = \frac{2}{20} = \frac{1}{10}$

$x_{k} = a + k ∆ x = - 1 + \frac{2 k}{10} x_{k - 1} = a + (k - 1) ∆ x = - 1 + \frac{2 (k - 1)}{10} = \frac{- 1 + 2 k - 2}{10} = \frac{2 k - 3}{10}$

and $\frac{x_{k - 1} + x_{k}}{2} = \frac{\frac{2 k - 3}{10} + \frac{2 k - 10}{10}}{2} = \frac{4 k - 13}{20}$

Step 3. Finding Reimann Sum

We have $f (x) = \sqrt{1 - x^{2}}$

Riemann sum

= \sum_{k = 1}^{20} f (\frac{x_{k - 1} + x_{k}}{2}) ∆ x = \sum_{k = 1}^{20} f (\frac{4 k - 13}{20}) \frac{1}{10} = \frac{1}{10} \sum_{k = 1}^{20} f (\frac{4 k - 13}{20}) = \frac{1}{10} \sum_{k = 1}^{20} \sqrt{1 - {(\frac{4 k - 13}{20})}^{2}}

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Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding ∆x and xk-1

Step 3. Finding Reimann Sum

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In Exercises 39–44, write out the sigma notation for the Riemann sum described in such a way that the only letter which appears in the general term of the sum is k. Don’t calculate the value of the sum; just write it down in sigma notation.fx=1-x2,a,b=-1,1,midpointsum,n=20

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding ∆x and xk-1

Step 3. Finding Reimann Sum

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Probability and Statistics

Logic and Functions

Calculus

Geometry

Statistics

Study anywhere. Anytime. Across all devices.