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

[Petrus]

Calculate the length of the curve

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

 

[earboth]

Calculate the length of the curve

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

I would do these calculations a little bit different:

$$\sqrt{1+\left(\frac x{36} - \frac9x \right)^2} = $$
$$ \sqrt{1+\frac{x^2}{36^2} - \frac{1}{2} + \frac{81}{x^2} }= $$
$$\sqrt{\left(\frac {x}{36} + \frac{9}{x} \right)^2} = \left| \frac {x}{36} + \frac{9}{x} \right|$$

 

[Petrus]

I would do these calculations a little bit different:

$$\sqrt{1+\left(\frac x{36} - \frac9x \right)^2} = $$
$$ \sqrt{1+\frac{x^2}{36^2} - \frac{1}{2} + \frac{81}{x^2} }= $$
$$\sqrt{\left(\frac {x}{36} + \frac{9}{x} \right)^2} = \left| \frac {x}{36} + \frac{9}{x} \right|$$

ehmm... Cant I do it my way?
edit: I have never done integrate with absolute value, how does that work?

 

[earboth]

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:

$$\int_9^{9e}\left(\frac x{36} + \frac9x \right) dx$$

 

[Petrus]

I would do these calculations a little bit different:

$$\sqrt{1+\left(\frac x{36} - \frac9x \right)^2} = $$
$$ \sqrt{1+\frac{x^2}{36^2} - \frac{1}{2} + \frac{81}{x^2} }= $$
$$\sqrt{\left(\frac {x}{36} + \frac{9}{x} \right)^2} = \left| \frac {x}{36} + \frac{9}{x} \right|$$

How do you cancel out 1? should it not be $$-\frac{1}{4}$$ insted of $$-\frac{1}{2}$$. I don't understand this part with integrate with absolute value can just cancel out?

 

[earboth]

How do you cancel out 1? should it not be $$-\frac{1}{4}$$ insted of $$-\frac{1}{2}$$. I don't understand this part with integrate with absolute value can just cancel out?

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

... and $$1-\frac12 = +\frac12$$

 

[MarkFL]

Calculate the length of the curve

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

 

[Petrus]

Hello,
I am still confused... Cant I just integrate with just subsitute?

 

[MarkFL]

I would just integrate term by term:

$$s=\frac{1}{36}\int_9^{9e}x\,dx+9\int_9^{9e}\frac{1}{x}\,dx$$

 

[Petrus]

Thanks for the help MarkFL and earboth!:)