1. Let's state the problem: We want to understand which Riemann sum, left or right, generally overestimates the integral of a function. 2. The formula for the left Riemann sum over an interval $[a,b]$ with $n$ subintervals is: $$L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x$$ where $x_i = a + i\Delta x$ and $\Delta x = \frac{b-a}{n}$. 3. The formula for the right Riemann sum is: $$R_n = \sum_{i=1}^n f(x_i) \Delta x$$ where $x_i = a + i\Delta x$. 4. Important rule: The behavior depends on whether the function $f$ is increasing or decreasing. 5. If $f$ is increasing on $[a,b]$, then: - The left sum $L_n$ underestimates the integral. - The right sum $R_n$ overestimates the integral. 6. If $f$ is decreasing on $[a,b]$, then: - The left sum $L_n$ overestimates the integral. - The right sum $R_n$ underestimates the integral. 7. Explanation: For an increasing function, the left endpoint values are always less than or equal to the function values on the subinterval, so the left sum is less than the true area. The right endpoint values are greater, so the right sum is greater than the true area. 8. For a decreasing function, the opposite happens: left endpoints are higher, so left sum overestimates; right endpoints are lower, so right sum underestimates. 9. Therefore, which sum overestimates depends on the monotonicity of the function. Final answer: Generally, for increasing functions, the right sum overestimates; for decreasing functions, the left sum overestimates.