How to calculate the arc length of a function using integration by parts?

In summary, the Calculate the length of the curve conversation reveals that:-The length of the curve can be calculated using the formula \int_a^b\sqrt{1+[f'(x)]^2}-This formula can be rewritten as \int_9^{9e}\sqrt{1+\frac{(x^2-324)^2}{1296x^2}}-The function can be integrated by part to get \int_9^{9e}\sqrt{\frac{1298x^2+(x^2-324)^2}{1296x^2}}-The resulting integral can be solved for the length of the
  • #1
Petrus
702
0
Calculate the length of the curve
1a53fc38d3d0ea39b31947e4b9dc841.png

We got the formula \(\displaystyle \int_a^b\sqrt{1+[f'(x)]^2}\)
and \(\displaystyle f'(x)=\frac{x}{36}-\frac{9}{x} <=> \frac{x^2-324}{36x}\)
so now we got \(\displaystyle \int_9^{9e}\sqrt{1+(\frac{x^2-324}{36x})^2}\)
we can rewrite that as \(\displaystyle \int_9^{9e}\sqrt{1+\frac{(x^2-324)^2}{1296x^2}}\)
then do integration by part on it well I simply rewrite the function more too \(\displaystyle \int_9^{9e}\sqrt{\frac{1298x^2+(x^2-324)^2}{1296x^2}}\) and set \(\displaystyle u=1298x^2+(x^2-324)^2\) so we got \(\displaystyle dv=1296x^{-2}\) so our \(\displaystyle du=2592x+2x^2-648+x\) and \(\displaystyle v=\frac{1296x^{-3}}{-3}\)Is this correct?
 
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  • #2
Petrus said:
Calculate the length of the curve
1a53fc38d3d0ea39b31947e4b9dc841.png

We got the formula \(\displaystyle \int_a^b\sqrt{1+[f'(x)]^2}\)
and \(\displaystyle f'(x)=\frac{x}{36}-\frac{9}{x} <=> \frac{x^2-324}{36x}\)
so now we got \(\displaystyle \int_9^{9e}\sqrt{1+(\frac{x^2-324}{36x})^2}\)
we can rewrite that as \(\displaystyle \int_9^{9e}\sqrt{1+\frac{(x^2-324)^2}{1296x^2}}\)
then do integration by part on it well I simply rewrite the function more too \(\displaystyle \int_9^{9e}\sqrt{\frac{1298x^2+(x^2-324)^2}{1296x^2}}\) and set \(\displaystyle u=1298x^2+(x^2-324)^2\) so we got \(\displaystyle dv=1296x^{-2}\) so our \(\displaystyle du=2592x+2x^2-648+x\) and \(\displaystyle v=\frac{1296x^{-3}}{-3}\)Is this correct?

I would do these calculations a little bit different:

\(\displaystyle \sqrt{1+\left(\frac x{36} - \frac9x \right)^2} = \)
\(\displaystyle \sqrt{1+\frac{x^2}{36^2} - \frac{1}{2} + \frac{81}{x^2} }= \)
\(\displaystyle \sqrt{\left(\frac {x}{36} + \frac{9}{x} \right)^2} = \left| \frac {x}{36} + \frac{9}{x} \right|\)
 
  • #3
earboth said:
I would do these calculations a little bit different:

\(\displaystyle \sqrt{1+\left(\frac x{36} - \frac9x \right)^2} = \)
\(\displaystyle \sqrt{1+\frac{x^2}{36^2} - \frac{1}{2} + \frac{81}{x^2} }= \)
\(\displaystyle \sqrt{\left(\frac {x}{36} + \frac{9}{x} \right)^2} = \left| \frac {x}{36} + \frac{9}{x} \right|\)
ehmm... Cant I do it my way?
edit: I have never done integrate with absolute value, how does that work?
 
Last edited:
  • #4
Petrus said:
ehmm... Cant I do it my way?
edit: I have never done integrate with absolute value, how does that work?
With the given domain your integral becomes:

\(\displaystyle \int_9^{9e}\left(\frac x{36} + \frac9x \right) dx\)
 
  • #5
earboth said:
I would do these calculations a little bit different:

\(\displaystyle \sqrt{1+\left(\frac x{36} - \frac9x \right)^2} = \)
\(\displaystyle \sqrt{1+\frac{x^2}{36^2} - \frac{1}{2} + \frac{81}{x^2} }= \)
\(\displaystyle \sqrt{\left(\frac {x}{36} + \frac{9}{x} \right)^2} = \left| \frac {x}{36} + \frac{9}{x} \right|\)
How do you cancel out 1? should it not be \(\displaystyle -\frac{1}{4}\) insted of \(\displaystyle -\frac{1}{2}\). I don't understand this part with integrate with absolute value can just cancel out?
 
  • #6
Petrus said:
How do you cancel out 1? should it not be \(\displaystyle -\frac{1}{4}\) insted of \(\displaystyle -\frac{1}{2}\). I don't understand this part with integrate with absolute value can just cancel out?

\(\displaystyle \left(\frac x{36} - \frac9x \right)^2 = \frac{x^2}{36^2} - \underbrace{2\cdot \frac x{36} \cdot \frac9x}_{\frac12} + \frac{9^2}{x^2}\)

... and \(\displaystyle 1-\frac12 = +\frac12\)
 
  • #7
Petrus said:
Calculate the length of the curve
1a53fc38d3d0ea39b31947e4b9dc841.png

We got the formula \(\displaystyle \int_a^b\sqrt{1+[f'(x)]^2}\)
and \(\displaystyle f'(x)=\frac{x}{36}-\frac{9}{x} <=> \frac{x^2-324}{36x}\)
so now we got \(\displaystyle \int_9^{9e}\sqrt{1+(\frac{x^2-324}{36x})^2}\)
we can rewrite that as \(\displaystyle \int_9^{9e}\sqrt{1+\frac{(x^2-324)^2}{1296x^2}}\)
then do integration by part on it well I simply rewrite the function more too \(\displaystyle \int_9^{9e}\sqrt{\frac{1298x^2+(x^2-324)^2}{1296x^2}}\) and set \(\displaystyle u=1298x^2+(x^2-324)^2\) so we got \(\displaystyle dv=1296x^{-2}\) so our \(\displaystyle du=2592x+2x^2-648+x\) and \(\displaystyle v=\frac{1296x^{-3}}{-3}\)Is this correct?

Petrus, just a minor quibble...when you write integrals, you should include the differential at the end that indicates with respect to what variable you are integrating. (Happy)
 
  • #8
Hello,
I am still confused... Cant I just integrate with just subsitute?
 
  • #9
I would just integrate term by term:

\(\displaystyle s=\frac{1}{36}\int_9^{9e}x\,dx+9\int_9^{9e}\frac{1}{x}\,dx\)
 
  • #10
Thanks for the help MarkFL and earboth!:)
 

FAQ: How to calculate the arc length of a function using integration by parts?

1. What is the definition of arc length of a function?

The arc length of a function is the distance along the curve of the function between two points. It is also known as the length of the curve or the length of the arc.

2. How is the arc length of a function calculated?

The arc length of a function can be calculated using the formula: L = ∫√(1 + [f'(x)]^2) dx, where f'(x) is the derivative of the function.

3. Why is it important to calculate the arc length of a function?

The arc length of a function is important because it helps us understand the shape and behavior of the function. It can also be used in various applications such as physics, engineering, and geometry to calculate distances or areas.

4. Can the arc length of a function be negative?

No, the arc length of a function cannot be negative. It is always a positive value as it represents a distance.

5. Is there a way to approximate the arc length of a function?

Yes, there are different methods to approximate the arc length of a function, such as using numerical integration techniques like the Trapezoidal Rule or Simpson's Rule. These methods can provide a close approximation of the exact arc length.

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