Question

find arc length: y= x^4/8 +1/4x^2 . 12

Answer 1

The integral that needs to be found is \[\int\limits_{1}^{2}\sqrt{1+(x^3/2-1/2x^3)}dx\]. Simplify the integrand first before evaluating it.

Answer 2

\[
1 + y'^2 =\frac{\left(x^6+1\right)^2}{4
x^6}\\

\sqrt{1 + y'^2} =\frac{\left(x^6+1\right)}{2 x^3}=
\\
\frac 1 2x^2 + \frac 1 {2 x^3}
\]
Integrate from 1 to 2 now

Answer 3

i dont get it... im getting different stuff: