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Setting up the midpoint Riemann sum limit for the area under a curve

Consider the function f(x)=3x−x2f(x) = 3x - x^2 on the interval [0,3][0,3].

The region under f(x)=3x−x2f(x) = 3x - x^2 on [0,3][0,3] lies entirely on or above the xx-axis, so its area equals ∫03(3x−x2) dx\displaystyle\int_0^3 (3x-x^2)\,dx.

Divide [0,3][0,3] into nn equal subintervals, each of width Δx\Delta x, and let xi∗x_i^* denote the midpoint of the ii-th subinterval. Write down the expressions for Δx\Delta x and xi∗x_i^* (in terms of nn and ii), and then express the area as a limit of a midpoint Riemann sum: ∫03(3x−x2) dx=lim⁡n→∞∑i=1nf(xi∗) Δx.\int_0^3 (3x-x^2)\,dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\,\Delta x.

State your final expression for lim⁡n→∞∑i=1nf(xi∗) Δx\displaystyle\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\,\Delta x with Δx\Delta x and xi∗x_i^* written explicitly in terms of nn and ii, and f(xi∗)f(x_i^*) substituted in using the formula for ff.

Hint

Since the interval [0,3][0,3] has length 3, Δx=3/n\Delta x = 3/n. The left endpoint of subinterval ii is 0+(i−1)Δx0 + (i-1)\Delta x, so the midpoint is that value plus half of Δx\Delta x. Substitute this midpoint into f(x)=3x−x2f(x) = 3x - x^2 to get f(xi∗)f(x_i^*).

Step-by-step solution

Since [0,3][0,3] is split into nn equal pieces, Δx=3−0n=3n\Delta x = \dfrac{3-0}{n} = \dfrac{3}{n}. The ii-th subinterval is [0+(i−1)Δx, 0+iΔx][0+(i-1)\Delta x,\ 0+i\Delta x], so its midpoint is xi∗=(i−1)Δx+Δx2=(i−12)Δx=3(2i−1)2nx_i^* = (i-1)\Delta x + \dfrac{\Delta x}{2} = \left(i - \dfrac{1}{2}\right)\Delta x = \dfrac{3(2i-1)}{2n}.

Substituting Δx=3/n\Delta x = 3/n and xi∗=3(2i−1)2nx_i^* = \dfrac{3(2i-1)}{2n} into f(xi∗)=3xi∗−(xi∗)2f(x_i^*) = 3x_i^* - (x_i^*)^2 and multiplying by Δx\Delta x gives the Riemann sum term. Taking the limit as n→∞n\to\infty, the area is lim⁡n→∞∑i=1n[3⋅3(2i−1)2n−(3(2i−1)2n)2]⋅3n.\lim_{n\to\infty}\sum_{i=1}^n \left[3\cdot\frac{3(2i-1)}{2n} - \left(\frac{3(2i-1)}{2n}\right)^2\right]\cdot\frac{3}{n}. This limit, once evaluated, equals the exact area under the curve.

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