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How do you set up an integral from the length of the curve \[y=\dfrac{1}{x},1\le x\le 5\] ?

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Last updated date: 13th Jun 2024
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Answer
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Hint: These types of problems are pretty straight forward and are very easy to solve. For solving such problems, we need to keep in mind the chapter Application of Derivatives and the given problem is a great example of it. We need to remember the main formula for finding the length of a curve. Suppose we have a curve \[y=f\left( x \right)\] , and we want to find the length of this curve from \[x=a\] and \[x=b\] , then the length of this curve in the given boundary is evaluated by the formula,
\[Length=\int\limits_{a}^{b}{\sqrt{1+{{\left( \dfrac{dy}{dx} \right)}^{2}}}dx}\]
Here \[\dfrac{dy}{dx}\] is the derivative of the given function.

Complete step by step solution:
Now, we start off with the solution by writing that,
\[Length=\int\limits_{a}^{b}{\sqrt{1+{{\left( \dfrac{dy}{dx} \right)}^{2}}}dx}\] . We find the value of the derivative of the function and then we plug that in the given formula for the length to find the required answer.
We have,
\[y=\dfrac{1}{x}\] . Differentiating, we get,
\[\dfrac{dy}{dx}=-\dfrac{1}{{{x}^{2}}}\]
The ‘a’ and ‘b’ in the formula is basically the values $1$ and $5$ respectively in this given problem. Thus we can represent this given problem, in terms of the formula to find out the length of the given curve between the given points as,
\[Length=\int\limits_{1}^{5}{\sqrt{1+{{\left( -\dfrac{1}{{{x}^{2}}} \right)}^{2}}}dx}\]
Evaluating we get,
\[Length=\int\limits_{1}^{5}{\sqrt{1+\dfrac{1}{{{x}^{4}}}}dx}\]
We can now evaluate the integral to find out the actual value of the length of the curve.

Note:
For problems like these we need to keep in mind the general formula to calculate the length of any given curve between any two points, or else finding the required answer may be cumbersome. We also need to evaluate the value of the integral carefully. However the given formula is applicable for both bounded and unbounded curves to find the length of curves.