# arc length of a function

#### Petrus

##### Well-known member
Calculate the length of the curve

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?

#### earboth

##### Active member
Calculate the length of the curve

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:

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

#### Petrus

##### Well-known member
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?

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#### earboth

##### Active member
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:

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

#### Petrus

##### Well-known member
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 dont understand this part with integrate with absolute value can just cancel out?

#### earboth

##### Active member
How do you cancel out 1? should it not be $$\displaystyle -\frac{1}{4}$$ insted of $$\displaystyle -\frac{1}{2}$$. I dont 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 $$\displaystyle 1-\frac12 = +\frac12$$

#### MarkFL

Staff member
Calculate the length of the curve

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.

#### Petrus

##### Well-known member
Hello,
I am still confused... Cant I just integrate with just subsitute?

#### MarkFL

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