## Calculus: Early Transcendentals 8th Edition

$\displaystyle \frac{32}{3}$ (converges)
The integrand has a discontinuity at -2, so this is a type 2.b improper integral. $\displaystyle \int_{a}^{b}f(x)dx=\lim_{t\rightarrow a^{+}}\int_{t}^{b}f(x)dx$ $I=\displaystyle \int_{-2}^{14}\frac{dx}{\sqrt[4]{x+2}}=\lim_{t\rightarrow-2^{+}}\int_{t}^{14}(x+2)^{-1/4}dx \\ \left[\begin{array}{ll} u=x+2 & \\ du=dx & \\ x=-2\Rightarrow u=0, & x=14\Rightarrow u=16 \end{array}\right]$ $=\displaystyle \lim_{u\rightarrow 0^{+}}\left[\frac{4}{3}u^{3/4}\right]_{u}^{16}$ $=\displaystyle \frac{4}{3}\left[16^{3/4}-0\right]$ $=\displaystyle \frac{4}{3}(2^{3})$ $=\displaystyle \frac{32}{3}$ (converges)