Question

y = (1/4)x^2 - (1/2)lnx..over the interval (1, 7e) ...what is the arc length ? need help pleease

Answer 1

s= integral from a to b sqrt(1+(y prime)^2)dx

note y prime is the derivative of your function with respect to x and your integral goes from 1 to 7e

Answer 2

or do you need to know how to do the derivatives and integrals?

Answer 3

The arclength is given by:
\[\int\limits_a^b \sqrt{1+[f'(x)]^2}dx\]
So you have:
\[f'(x)=(1/2)x-1/(2x) \rightarrow [f'(x)]^2=((1/2)x-1/(2x))^2=((1/4)x^2+1/(4x^2)-(1/2))\]
Then you have:
\[\int\limits_1^{7e} \sqrt{1+(1/4)x^2+1/(4x^2)+(1/2)}dx\]
As long as my mental squaring is correct.

Answer 4

i have ((x^2)/4)- ((x^2)/4)+(1/(4x^2))

Answer 5

that was for my derivative part ..//

Answer 6

that was my deravitive part that i squared and simplified...

Answer 7

http://www.wolframalpha.com/input/?i=integrate+sqrt((3%2F2)%2B(1%2F4)x^2%2B1%2F(4x^2))