Midpoint Riemann sum for f(x) = sqrt(x)

With $a=1$, $b=9$, and $n=4$, the width of each subinterval is $\Delta x = \frac{9-1}{4} = 2$. This gives subintervals $[1,3], [3,5], [5,7], [7,9]$, with midpoints $2, 4, 6, 8$.

Evaluate $f$ at each midpoint: $f(2)=\sqrt{2}$, $f(4)=\sqrt{4}=2$, $f(6)=\sqrt{6}$, $f(8)=\sqrt{8}=2\sqrt{2}$. The midpoint sum is $M_4 = \Delta x \left[f(2)+f(4)+f(6)+f(8)\right] = 2\left(\sqrt{2}+2+\sqrt{6}+2\sqrt{2}\right)$.

Computing numerically: $\sqrt{2}\approx 1.4142$, $\sqrt{6}\approx 2.4495$, $2\sqrt{2}\approx 2.8284$. So the sum inside the brackets is approximately $1.4142+2+2.4495+2.8284 = 8.6921$, and $M_4 = 2(8.6921) \approx 17.384$.

Since $f''(x) = -\frac{1}{4x^{3/2}} < 0$, the function $\sqrt{x}$ is concave down on $[1, 9]$. For concave-down functions, the midpoint rule underestimates the true area under the curve. This is because at each midpoint, the curve lies above the top of the rectangle, so the rectangular area is smaller than the true area under the curve. Therefore, the answer is underestimate.