Title: Integraali

Määritä käyrän

\[ y=\frac{x^3}{6}+\frac{1}{2x} \]

pituus välillä \([value]\leq x\leq [value]\). Anna vastaus tarkkana arvona!

Vastaus:
Feedback: Mallivastaus: Käyrän \( y=f(x)=\frac{x^3}{6}+\frac{1}{2x} \) pituus \(S\) saadaan laskettua kaavalla \[S= \int_{a}^{b} \sqrt{1+[f'(x)]^2} \, \mathrm{d}x. \] Rajat ovat \([value]\) ja \([value]\). \[f'(x)=\frac{1}{2} x^2-\frac{1}{2} x^{-2} \] \begin{eqnarray*} S & = & \int_{[value]}^{[value]} \sqrt{1+\frac{1}{4}(x^2-x^{-2})^2} \, \textrm{d}x \\ & = & \int_{[value]}^{[value]} \sqrt{1+\frac{1}{4}(x^4-2x^0+x^{-4})} \, \textrm{d}x \\ & = & \int_{[value]}^{[value]} \sqrt{1+\frac{1}{4}x^4-\frac{1}{2}+\frac{1}{4}x^{-4}} \, \textrm{d}x \\ & = & \frac{1}{2} \int_{[value]}^{[value]} \sqrt{x^4+2+x^{-4}} \, \textrm{d}x \\ & = & \frac{1}{2} \int_{[value]}^{[value]} \sqrt{(x^2+x^{-2})^2} \, \textrm{d}x \\ & = & \frac{1}{2} \int_{[value]}^{[value]} (x^2+x^{-2}) \, \textrm{d}x \\ & = & \frac{1}{2} \bigg|_{[value]}^{[value]} \frac{1}{3} x^3-x^{-1} \\ & = & \frac{1}{2}((\frac{[value]}{3}-\frac{1}{[value]})-(\frac{[value]}{3}-\frac{1}{[value]})) \\ & = & \frac{1}{2}([value]-[value]) \\ & = & [value]. \end{eqnarray*}