ineedhelpnow said:
$y^2=4(x+4)^3$
$0 \le x \le 2$
$y=2(x+4)^{3/2}$
$y'=3(x+4)^{1/2}$
$\int_{0}^{2} \ \sqrt{1+9(x+4)},dx = 13.5429$
is that right?
Your setup of the integral is correct. You can probably get an exact answer though.
$\displaystyle \begin{align*} \int_0^2{ \sqrt{ 1 + 9 \left( x + 4 \right) } \,\mathrm{d}x } &= \int_0^2{ \sqrt{ 1 + 9x + 36 }\,\mathrm{d}x } \\ &= \int_0^2{ \sqrt{ 9x + 37 } \, \mathrm{d}x } \\ &= \frac{1}{9} \int_0^2{ 9\,\sqrt{ 9x + 37 } \, \mathrm{d}x } \end{align*}$
and now make the substitution $\displaystyle \begin{align*} u = 9x + 37 \implies \mathrm{d}u = 9\,\mathrm{d}x \end{align*}$ and noting that $\displaystyle \begin{align*} u(0) = 37 \end{align*}$ and $\displaystyle \begin{align*} u(2) = 55 \end{align*}$, the integral becomes
$\displaystyle \begin{align*} \frac{1}{9} \int_0^2{ 9\,\sqrt{ 9x + 37} \, \mathrm{d}x } &= \frac{1}{9} \int_{37}^{55}{ \sqrt{u}\,\mathrm{d}u } \\ &= \frac{1}{9} \int_{37}^{55}{ u^{\frac{1}{2}}\,\mathrm{d}u } \\ &= \frac{1}{9} \left[ \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right] _{37}^{55} \\ &= \frac{1}{9} \left[ \frac{2}{3} u\,\sqrt{u} \right] _{37}^{55} \\ &= \frac{2}{27} \left( 55\,\sqrt{55} - 37\,\sqrt{37} \right) \end{align*}$
