Integrais

Seja f(x)= (x⁴/ + (1/4x²), o valor da integral √[1+(f'(x))²] no intervalo (1,2) é:

Gabarito: 33/16

\[
f'(x) = \frac{x^3}{2} + \frac{1}{4} \cdot (-2) \cdot \frac{1}{x^3} = \frac{x^3}{2} - \frac{1}{2x^3} = \frac{1}{2} \left( x^3 - \frac{1}{x^3} \right)
\]
\[
[f'(x)]^2 = \frac{1}{4} \left( x^6 + \frac{1}{x^6} - 2 \right)
\]
\[\begin{aligned}
I &= \int_1^2 \sqrt{ 1 + \frac{1}{4} \left( x^6 + \frac{1}{x^6} - 2 \right) } \ \mathrm{d} x \\
& = \frac{1}{2} \cdot \int_1^2 \sqrt{4 + x^6 -2 + \frac{1}{x^6} } \ \mathrm{d} x \\
& = \frac{1}{2} \cdot \int_1^2 \sqrt{ 2 + x^6 + \frac{1}{x^6}} \ \mathrm{d}x \\
& = \frac{1}{2} \cdot \int_1^2 \sqrt{ \left( x^3 + x^{-3} \right)^2 } \ \mathrm{d} x \\
& = \frac{1}{2} \cdot \int_1^2 \left| x^3 + x^{-3} \right| \ \mathrm{d} x \\
& = \frac{1}{2} \cdot \int_1^2 x^3 + x^{-3} \ \mathrm{d}x\\
& = \frac{1}{2} \left( \frac{x^4}{4} - \frac{x^{-2}}{2} \right)\Big|_1^2 \\
& = \frac{1}{2} \left( \frac{16}{4} - \frac{1}{8} - \frac{1}{4} + \frac{1}{2} \right) \\
& = \frac{1}{2} \cdot \frac{33}{8} \\
& = \frac{33}{16}
\end{aligned}
\]