University:
Whitman College
Course:
MATH-126 | Calculus II
Academic year:

2024
Views:

156

Pages:

2
Author:

suddsscatheless

Solutions to Extra Practice in Riemann Sums 1. For each of the following integrals, write the definition using the Riemann sum (and right endpoints), but do not evaluate them: (a) Z 5 2 (b) sin(3x) dx = n→∞ lim Z 3√ 1 + x dx = lim n→∞ 1 (c) Z 2 n X ex dx = lim n→∞ 0 n X sin (3(2 + 3i/n)) i=1 n q X 1 + (1 + 2i/n) i=1 e2i/n i=1 3 n 2 n 2 n 2. For each of the following integrals, write the definition using the Riemann sum, and then evaluate them (MUST use the limit of the Riemann sum for credit, and do not re-write them using the properties of the integral): (a) Z 5 x2 dx 2 lim n→∞ n X 3i 2+ n i=1 2 n 3 3X 12i 9i2 = n→∞ lim 4+ + 2 n n i=1 n n ! = n n n X 12i X 9i2 3 X 4+ + = lim 2 n→∞ n i=1 n i=1 n i=1 " # n n n 3 X 12 X 9 X lim 4 1 + i + i2 = 2 n→∞ n n n i=1 i=1 i=1 " # " # 12 n(n + 1) 9 n(n + 1)(2n + 1 3 4n + + 2 = lim n→∞ n n 2 n 6 lim n→∞ (b) Z 3 n + 1 9 (n + 1)(2n + 1) 12 + 18 · + · n 2 n2 ! = 12 + 18 + 9 = 39 1 − 3x dx 1 lim n→∞ (c) Z 5 n X i=1 2i 1−3 1+ n n X 2 6i 2 2 = lim −2 − = lim − (2n+3(n+1)) = −10 n→∞ n→∞ n n n n i=1 1 + 2x3 dx 0 With (b − a)/n = 5/n, the Riemann sum is given by: n n n 5X 53 i3 5 X 2 · 53 X lim 1 + 2 3 = lim 1+ 3 i3 = n→∞ n n→∞ n n n i=1 i=1 i=1 # " 5 2 · 53 n2 (n + 1)2 54 (n + 1)2 54 lim n + = lim 5 + = 5 + n→∞ n n→∞ n3 22 2 n2 2 " # " Without a calculator, it’s fine to leave it in that form. 1 # 3. For each of the following Riemann sums, evaluate the limit by first recognizing it as an appropriate integral: (a) n→∞ lim n X 3 s n Some options: 1+ i=1 Z 3√ 1 + x dx 3i (Find four different integrals for this one!) n Z 4q 0 1 + (x − 1) dx = 1 n X 25i2 (b) lim 2+3· 2 n→∞ n i=1 Some options: Z 5 1 + (x − 2) dx = 2 ! 5 n 2 + 3x2 dx 2i sin 3 + (c) n→∞ lim n i=1 Some options: Z 5q x dx 1 0 n X Z 4√ Z 6 2 + 3(x − 1)2 dx 1 Z 7 2 + 3(x − 2)2 dx 2 2 n Z 2 sin(3 + x) dx Z 5 3 0 2 sin(x) dx etc. Z 5√ 2 x−1

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