# How do you use Riemann sums to evaluate the area under the curve of y = x^2 + 1 on the closed interval [0,1], with n=4 rectangles using midpoint?

$\frac{43}{32}$
${\sum}_{n = 1}^{4} f \left(\frac{2 n - 1}{8}\right) \left(\frac{1}{4}\right) = \frac{1}{4} \left(\frac{1}{64} + 1 + \frac{9}{64} + 1 + \frac{25}{64} + 1 + \frac{49}{64} + 1\right) = \frac{1}{4} \left(4 + \frac{84}{64}\right) = 1 + \frac{22}{64} = 1 + \frac{11}{32} = \frac{43}{32}$