Right-endpoint Riemann sum vs. exact integral

With $\Delta x = (5-1)/4 = 1$, the right endpoints are $x_1=2, x_2=3, x_3=4, x_4=5$. Evaluating $f(x)=x^2-4x+5$ gives $f(2)=1$, $f(3)=2$, $f(4)=5$, $f(5)=10$. The right-endpoint sum is $\Delta x [f(2)+f(3)+f(4)+f(5)] = 1(1+2+5+10) = 18$.

An antiderivative of $f$ is $F(x) = \frac{x^3}{3} - 2x^2 + 5x$. Then $F(5) = \frac{125}{3} - 2(25) + 5(5) = \frac{125}{3} - 50 + 25 = \frac{125}{3} - 25 = \frac{50}{3}$ and $F(1) = \frac{1}{3} - 2(1) + 5(1) = \frac{1}{3} - 2 + 5 = \frac{10}{3}$. Thus $\int_1^5 f(x)\,dx = F(5) - F(1) = \frac{50}{3} - \frac{10}{3} = \frac{40}{3} \approx 13.33$.

Since the right sum (18) is greater than the exact value ($40/3 \approx 13.33$), the right-endpoint sum overestimates the integral. This is because $f(x) = x^2 - 4x + 5$ has $f'(x) = 2x - 4$. On the subinterval $[1,2]$, $f'(x) < 0$, so $f$ is decreasing and the right endpoint gives a value below the average on that interval, resulting in an underestimate for that subinterval. On the three subintervals $[2,3]$, $[3,4]$, and $[4,5]$, $f'(x) > 0$, so $f$ is increasing and the right endpoints give values above the average on each interval, resulting in overestimates for those three subintervals. Since the function grows more steeply as $x$ increases (the parabola opens upward), the three overcounts on the increasing subintervals outweigh the one undercount on the decreasing subinterval, producing a net overestimate.