Q.60 Prove Theorem 4.13(b): For any r... [FREE SOLUTION]

Chapter 4: Q.60 (page 354)

Prove Theorem 4.13(b): For any real numbers a and b, we have ${鈭�}_{a}^{b} x d x = \frac{1}{2} (b^{2} - a^{2})$ . Use the proof of Theorem 4.13(a) as a guide.

Short Answer

Expert verified

For any real number a and b, ${鈭�}_{a}^{b} x d x = \frac{1}{2} (b^{2} - a^{2})$ is verified as,

${鈭�}_{a}^{b} x d x = \lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{n} [a + k (\frac{b - a}{n})] (\frac{b - a}{n}) = \lim_{n 鈫� 鈭�} a n (\frac{b - a}{n}) + \lim_{n 鈫� 鈭�} {(\frac{b - a}{n})}^{2} [\frac{n (n + 1)}{2}] = \frac{1}{2} (b^{2} - a^{2})$

Step by step solution

Step 1. Given information

The given Integral is ${鈭�}_{a}^{b} x d x = \frac{1}{2} (b^{2} - a^{2}) .$

Step 2. Proof

Take the interval $[a, b]$

$鈭� x = \frac{b - a}{n} x_{k} = a + k 鈭� x x_{k} = a + k (\frac{b - a}{n})$

Use the definition of definite integral to find ${鈭�}_{a}^{b} x d x .$

role="math" localid="1648579869591" ${鈭�}_{a}^{b} x d x = \lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{n} f (\overset{*}{x_{k}}) 鈭� x = \lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{n} [a + k (\frac{b - a}{n})] (\frac{b - a}{n}) = \lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{n} a (\frac{b - a}{n}) + \lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{n} k {(\frac{b - a}{n})}^{2} = \lim_{n 鈫� 鈭�} a n (\frac{b - a}{n}) + \lim_{n 鈫� 鈭�} {(\frac{b - a}{n})}^{2} [\frac{n (n + 1)}{2}] = \frac{1}{2} (b^{2} - a^{2})$

so ${鈭�}_{a}^{b} x d x = \frac{1}{2} (b^{2} - a^{2})$ for any real number a and b.

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Prove Theorem 4.13(b): For any real numbers a and b, we have ${鈭�}_{a}^{b} x d x = \frac{1}{2} (b^{2} - a^{2})$ . Use the proof of Theorem 4.13(a) as a guide.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Proof

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Prove Theorem 4.13(b): For any real numbers a and b, we have鈭�abxdx=12b2-a2. Use the proof of Theorem 4.13(a) as a guide.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Proof

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Prove Theorem 4.13(b): For any real numbers a and b, we have ${鈭�}_{a}^{b} x d x = \frac{1}{2} (b^{2} - a^{2})$ . Use the proof of Theorem 4.13(a) as a guide.