PART A: Express the area under the curve y = 9 − x^2 from 0 to 3 as a limit, using right endpoints. PART B: Find the exact area by evaluating the limit in part (a).

Answer 1

a = 0 , b = 3 , f(x) = 9 - x² , right-hand endpts

Imagine n as a finite number of rectangles of equal width, with bases on the x-axis, with heights = to the y-value of the function. For example, let n = 12. We have 12 rectangles beneath the concave down parabola. The width of the entire interval, b - a, is 3. If we divide the interval into 12 subintervals of = width, then each rectangle is 1/4 wide (which is (b-a)/n). The righthand endpts of these subints are x = 1/4 , 2/4 , 3/4 , ... , 12/4.

The heights of these rectangles would be f(1/4) , f(2/4) , ... , and f(12/4). The sum of the areas of the rectangles would then be ∑_k=1¹² f(k/4)·(1/4) or , replacing f(x) with 9 - x² , ∑_k=1¹² [9 - (k/4)²]·(1/4)

Then, we generalize further by swapping in n for 12: ∑_k=1ⁿ [9 - (3k/n)²]·(3/n). If we slap lim_n→∞ in front, that's the exact area, aka the definite integral: ∫₀³(9 - x²)dx = lim_n→∞ ∑_k=1ⁿ [9 - (3k/n)²]·(3/n).

The next step is to square 3k/n and distribute the 3/n to get lim_n→∞ ∑_k=1ⁿ [ 27/n - 27k²/n³]. Splitting this into separate sigmas, and letting the n terms "pass through" the sigmas (they are independent of k), gives

lim_n→∞ [(1/n) · ∑_k=1ⁿ 27 - (1/n³) · ∑_k=1ⁿ 27k²]

To evaluate this limit, we need to recall some sigma rules and formulas and some limits of rational functions.

∑_k=1ⁿ a = an where a is a constant. ∑_k=1ⁿ k = ((n)(n+1))/2 , and ∑_k=1ⁿ k² = ((n)(n+1)(2n+1))/6 = (2n³+3n²+n)/6

lim_n→∞ (1/n) · (27n) - (1/n³) ·. (27)(2n³+3n²+n)/6) = lim_n→∞ 27 - (9n³ + ... + ...)/(n³) = 27 - 9 = 18.