University:
Amherst College
Course:
MATH 111 | Introduction to the Calculus
Academic year:

2024
Views:

318

Pages:

3
Author:

Bennett Wheeler

Definite Integral Limit Definition using Riemann Sums Definition: the Definite Integral of a function f from x = a to x = b is given by Z n X b f (x) dx = lim (•) n→∞ a f (xi )∆x i=1 = lim [f (x1 )∆x + f (x2 )∆x + f (x3 )∆x + . . . + f (xi )∆x + . . . + f (xn )∆x] n→∞ Note: The definite integral is a limit of a sum! Just think about this formula as the limiting value of the sum of the areas of finitely many (n) approximating rectangles. To compute definite integrals the long (limit) way, follow these steps : Z Step 1: Given the integral b f (x) dx, pick off or identify the integrand f (x), and limits of a integration a and b. b−a . This width of each partitioned interval should be in terms of n. n Step 2: Compute ∆x = Step 3: Compute xi = a + i∆x . Leave the i as your counter. You have the left-most endpoint a from Step 1. You have width ∆x from Step 2. This endpoint xi should be in terms of i and n. Step 4: Plug xi and ∆x into the formula (•) above. Here it is again: Z (•) b f (x) dx = lim n→∞ a n X f (xi )∆x ←− MEMORIZE! i=1 Step 5: Use the following formulas for sum of integers i and finish evaluating the limit in n. n X 1=n i=1 (∗) n X i= i=1 (∗∗) n X n(n + 1) 2 i2 = i=1 (∗ ∗ ∗) n X i=1 3 i = n(n + 1)(2n + 1) 6 n(n + 1) 2 2 Note: your final answer for the definite integral should be a number after you finish the limit. 1. Read through the entire next problem. Make sure you understand the formula to start, as well as the formulas for ∆x and xi . Because it doesn’t feel natural yet, just trust the formulas right now. 6 Z x2 dx using the Limit Definition of the Definite Integral using Riemann Sums. Evaluate 0 Here f (x) = Z x2 , b−a 6−0 6 a = 0, b = 6, ∆x = = = and xi = a+i∆x = 0+i n n n 6 2 x dx = lim 0 n→∞ n X i=1 6 6i = . n n n X 6i 6 f (xi )∆x = lim f n→∞ n n i=1 2 ! n X 6 6i = lim n→∞ n n i=1 n 6 X 36i2 n→∞ n n2 = lim i=1 n 216 X 2 i n3 = lim n→∞ i=1 216 n(n + 1)(2n + 1) · n3 6 216 n(n + 1)(2n + 1) · 6 n3 216 n(n + 1)(2n + 1) · 6 n·n·n 216 n n+1 2n + 1 · · · 6 n n n 216 1 1 ·1· 1+ · 2+ 6 n n = lim n→∞ = lim n→∞ = lim n→∞ = lim n→∞ = lim n→∞ = ! 216 216 ·1·2= = 72 6 3 using (∗∗) 2. Read through the entire next problem. Make sure you understand the formula to start, as well as the formulas for ∆x and xi . Here the lower limit of integration a is not 0. Z 4 6 − 3x dx using the Limit Definition of the Definite Integral using Riemann Evaluate 1 Sums. b−a 4−1 3 Here f (x) = 6 − 3x, a = 1, b = 4, ∆x = = = n n n 3 3i and xi = a + i∆x = 1 + i =1+ . n n Z 4 6 − 3x dx = lim 1 n→∞ n X i=1 n X 3i 3 f 1+ n→∞ n n f (xi )∆x = lim i=1 n X 3i 3 = lim 6−3 1+ n→∞ n n i=1 ! n 3X 9i 3− n n = lim n→∞ i=1 3 n = lim n→∞ n X i=1 3− n X 9i i=1 !! n ! n n 9X 27 X 1− 2 i n n = lim n→∞ i=1 = lim n→∞ i=1 9 27 n(n + 1) (n) − 2 n n 2 using (∗) 27 n n + 1 = lim 9 − n→∞ 2 n n 27 1 = lim 9 − (1) 1 + n→∞ 2 n =9− = − 9 2 27 2

Definite Integral Limit Definition Using Riemann Sums

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