$$\int_{\2}^{\9}(\ $\sqrt{\8+x^3}$ ) d\x $\$ $$

asked by guest
on Nov 18, 2024 at 6:50 am



You asked:

Evaluate the integral: \(\int_{2}^{9}{\sqrt{8 + {x}^{3}}}dx\)

MathBot Answer:

Definite integral

\[\int_{2}^{9}{\sqrt{8 + {x}^{3}}}dx = \frac{6 \sqrt{2} \Gamma\left(\frac{1}{3}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {\frac{729 e^{\pi i}}{8}} \right)}}{\Gamma\left(\frac{4}{3}\right)} - \frac{4 \sqrt{2} \Gamma\left(\frac{1}{3}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {-1} \right)}}{3 \Gamma\left(\frac{4}{3}\right)} \approx 97.76540398279852528315577245092258\]


\(i\) is the imaginary unit, defined as \(i^2 = -1\).

\(\Gamma\) is the gamma function, which is an extension of the factorial function to complex number arguments. \(\Gamma\) is defined for any positive integer \(n\) with the formula \(\Gamma(n) = (n-1)!\) and for complex numbers with a positive real part with the formula \(\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt\).