Question

I cannot find where I messed up in this problem.
Find the exact arc length of y=sqrt(x-x^2)+sin^-1(sqrt(x))

Answer 1

from 0 to 1 i take it

Answer 2

you need
\[\int_0^1\sqrt{1+f'(x)^2}dx\] if i recall correctly, so first you need
\[f'(x)=\frac{\sqrt{x-x^2}}{x}\]

Answer 3

and so
\[f'(x)^2=\frac{x-x^2}{x^2}=\frac{1}{x}-x\]

Answer 4

add 1 and get
\[\frac{-x^2+x+1}{x}\]

Answer 5

damn that was wrong
\[f'(x)^2=\frac{1}{x}-1\] so
\[1+f'(x)^2=\frac{1}{x}\]

Answer 6

now you have
\[\int_0^1\sqrt{\frac{1}{x}}dx\] which should be ok right?

Answer 7

no, there were not any intergration limits

Answer 8

It wanted the exact value

Answer 9

yea I got that

Answer 10

but when I put in 2sqrt(x) as my answer, it said it was wrong

Answer 11

well the funciton is only defined on the interval [0,1] so i assumed those were the limits of integration

Answer 12

oh okay, I just assumed it meant no limits

Answer 13

integral is 2

Answer 14

i guess when it said "exact arc length" it meant the total arc length

Answer 15

Haha, yay I got it. Thank for your help!!! Yea, I did not exactly understand that part. But it is all good now.

Answer 16

great!
yw