The integral for the surface area is,

$$S = \iint\limits_{D}{{\sqrt {{{\left[ {{x^3}} \right]}^2} + {{\left[ 2 \right]}^2} + 1} \,dA}} = \iint\limits_{D}{{\sqrt {{x^6} + 5} \,dA}}$$ Show Step 3

Now, as mentioned in Step 1 the region $D$ is shown in the sketches of the surface. Here is a 2D sketch of $D$ for the sake of completeness.

The limits for this region are,

$$\begin{array}{c}0 \le x \le 1\\ 0 \le y \le {x^5}\end{array}$$

Note as well that the integrand pretty much requires us to do the integration in this order.

Show Step 4

With the limits from Step 3 the integral becomes,

$$S = \iint\limits_{D}{{\sqrt {{x^6} + 5} \,dA}} = \int_{0}^{1}{{\int_{0}^{{{x^5}}}{{\sqrt {{x^6} + 5} \,dy}}\,dx}}$$ Show Step 5

Okay, all we need to do then is evaluate the integral.

$$\begin{align*}S & = \int_{0}^{1}{{\int_{0}^{{{x^5}}}{{\sqrt {{x^6} + 5} \,dy}}\,dx}}\\ & = \int_{0}^{1}{{\left. {\left( {y\sqrt {{x^6} + 5} } \right)} \right|_0^{{x^5}}\,dx}}\\ & = \int_{0}^{1}{{{x^5}\sqrt {{x^6} + 5} \,dx}}\\ & = \left. {\frac{1}{9}{{\left( {{x^6} + 5} \right)}^{\frac{3}{2}}}} \right|_0^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{9}\left( {{6^{\frac{3}{2}}} - {5^{\frac{3}{2}}}} \right) = 0.3907}}\end{align*}$$