Chapter 6: Q. 5 (page 538)

How is the Mean Value Theorem involved in proving that the arc length of a function on an interval can be represented by a definite integral?

Short Answer

Expert verified

Arc length is $\lim_{n \to \infty} \sum_{k = 1}^{n} \sqrt{1 + {(f' (x_{k}^{*}))}^{2}} ∆ x = \int_{a}^{b} \sqrt{1 + {(f' (x))}^{2}} d x$

Step by step solution

Step 1. Given Information.

The mean value theorem states that,

If $f (x)$ is differential and continuous at $[a, b]$ , then there is c such as $a < c < b$ and

$f' (c) = \frac{f (b) - f (a)}{b - a}$ .

Step 2. Explanation.

To find arc length using Mean value theorem.

If $f (x)$ is a differentiable function, then the Mean Value Theorem applies to the function on each subinterval $[x_{k - 1}, x_{k}]$ then, there exists some point $x_{k}^{*} \in (x_{k - 1}, x_{k})$

such that,

role="math" localid="1650611993068" $f' (x_{k}^{*}) = \frac{f (x_{k}) - f (x_{k - 1})}{x_{k} - x_{k - 1}} = \frac{∆ y_{k}}{∆ x}$

so,

role="math" localid="1650612030835" $\lim_{n \to \infty} \sum_{k = 1}^{n} \sqrt{1 + {(\frac{∆ y_{k}}{∆ x})}^{2}} ∆ x = \lim_{n \to \infty} \sum_{k = 1}^{n} \sqrt{1 + {(f' (x_{k}^{*}))}^{2}} ∆ x$

As the derivative $f' (x)$ is assumed to be continuous, then function is also continue.

therefore, the limit of sums represents the definite integral of $\sqrt{1 + {(f' (x))}^{2}}$ on interval $[a, b]$ .

so that,

$\lim_{n \to \infty} \sum_{k = 1}^{n} \sqrt{1 + {(f' (x_{k}^{*}))}^{2}} ∆ x = \int_{a}^{b} \sqrt{1 + {(f' (x))}^{2}} d x$

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How is the Mean Value Theorem involved in proving that the arc length of a function on an interval can be represented by a definite integral?

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Explanation.

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