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Chapter 4: Q. 44 (page 340)

In Exercises 39鈥�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鈥檛 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 $\frac{1}{10} {鈭�}_{k = 1}^{20} \sqrt{1 - {(\frac{4 k - 13}{20})}^{2}}$

Step by step solution

01

Step 1. Given Information

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

02

Step 2. Finding ∆x and xk-1

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

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

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

03

Step 3. Finding Reimann Sum

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

Riemann sum

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

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Most popular questions from this chapter

Compare the definitions of the definite and indefinite integrals. List at least three things that are different about these mathematical objects.

Use the Fundamental Theorem of Calculus to find the exact values of the given definite integrals.

Use a graph to check your answer.

${鈭�}_{鈭� 3}^{2} 鈥� \frac{1}{(x + 5)^{2}} d x$

Show that $F (x) = \frac{1}{2} e^{(x^{2})} (x^{2} - 1)$ is an anti-derivative of $f (x) = x^{3} e^{(x^{2})} .$

Consider the general sigma notation ${鈭�}_{k = m}^{n} a_{k}$ .What do we mean when we say that a_k is a function of k?

Given a simple proof that if n is a positive integer and c is any real number, then ${鈭�}_{k = 1}^{n} c = c n$

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