Answer
$\int_{0}^{1}~\sqrt{x^3+1}~dx \approx 1.1097$
Work Step by Step
$\Delta x = \frac{b-a}{n} = \frac{1-0}{5} = 0.2$
We can find the midpoints of the five subintervals:
$x_1 = 0.1$
$x_2 = 0.3$
$x_3 = 0.5$
$x_4 = 0.7$
$x_5 = 0.9$
$\int_{0}^{1}~\sqrt{x^3+1}~dx \approx \sum_{i=1}^{5} f(x_i)~\Delta x$
$\int_{0}^{1}~\sqrt{x^3+1}~dx \approx 0.2\cdot (\sqrt{0.1^3+1}+\sqrt{0.3^3+1}+\sqrt{0.5^3+1}+\sqrt{0.7^3+1}+\sqrt{0.9^3+1})$
$\int_{0}^{1}~\sqrt{x^3+1}~dx \approx 0.2\cdot (5.54836)$
$\int_{0}^{1}~\sqrt{x^3+1}~dx \approx 1.1097$
