Q. 46 For each definite integral in Ex... [FREE SOLUTION]

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Chapter 4: Q. 46 (page 353)

For each definite integral in Exercises $41 鈥� 46$ ,
(a) find the general n-rectangle right sum and simplify your answer with sum formulas. Then
(b) use your answer to approximate the definite integral with $n = 100$ and $n = 1000$ . Finally,
(c) take the limit as $n 鈫� 鈭�$ to find the exact value.
${鈭�}_{- 1}^{2} (1 - x^{2}) d x$

Short Answer

Expert verified

Part (a) The right sum is $\frac{9 (n + 1) (2 n + 1)}{2 n^{2}} - \frac{9 (n + 1)}{n}$

Part (b) The right sums are $0.00 & 0.00$

Part (c) The exact value is zero

Step by step solution

Step 1. Given information

An expression is given as ${鈭�}_{- 1}^{2} (1 - x^{2}) d x$

Part (a) Step 1. General n-rectangle sum

The limits given are $[a, b] = [- 1, 2]$

Therefore,

$螖 x = \frac{h - a}{n} = \frac{2 + 1}{n} = \frac{3}{n} x_{k} = a + k 螖 x = - 1 + k (3 / n) = - 1 + (3 k / n)$

Now the right sum is,

$= {鈭�}_{k = 1}^{n} f (x_{k}) 螖 x = {鈭�}_{k = 1}^{n} (1 - x_{k}^{2}) 螖 x = {鈭�}_{k = 1}^{n} (1 - (- 1 + (3 k / n))^{2}) (3 / n) = \frac{3}{n} {鈭�}_{k = 1}^{n} (1 - 1 + \frac{9 k^{2}}{n^{2}} - \frac{6 k}{n}) = \frac{3}{n} {鈭�}_{k = 1}^{n} (\frac{9 k^{2}}{n^{2}} - \frac{6 k}{n}) = \frac{3}{n} (- \frac{6 n (n + 1)}{2 n} + \frac{9 n (n + 1) (2 n + 1)}{6 n^{2}}) = \frac{9 (n + 1) (2 n + 1)}{2 n^{2}} - \frac{9 (n + 1)}{n}$

Part (b) Step 1. Approximate integral

The right sum is,

$\frac{9 (n + 1) (2 n + 1)}{2 n^{2}} - \frac{9 (n + 1)}{n} 鈬� n = 100 鈬� = 0.45 鈬� n = 1000 鈬� = 0.045$

For both values ofnwe get zero.

Part (c) Step 1. Exact value

Put the limits in the right sum as,

$= \frac{9 (n + 1) (2 n + 1)}{2 n^{2}} - \frac{9 (n + 1)}{n} \lim_{n 鈫� 鈭�} {鈭�}_{- 1}^{2} (1 - x^{2}) d x = \lim_{n 鈫� w} (\frac{9 (n + 1) (2 n + 1)}{2 n^{2}} - \frac{9 (n + 1)}{n}) = 9 - 9 = 0$

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Short Answer

Step by step solution

Step 1. Given information

Part (a) Step 1. General n-rectangle sum

Part (b) Step 1. Approximate integral

Part (c) Step 1. Exact value

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