- #1
Petrus
- 702
- 0
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?
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?