Math 111, Fall 2014 Worksheet 8

Math 111, Fall 2014 Worksheet 8, Tuesday, November 11th, 2014 Definition: the Definite Integral of 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: b Z f (x) dx, pick off or identify the integrand f (x), and limits of Step 1: Given the integral a integration a and b. Step 2: Compute ∆x = b−a . This width of each partitioned interval should be in terms of n. n Step 3: Compute xi = a + i∆x . Leave the i as your counter. You have a from Step 1. You have ∆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=1 n X i= n(n + 1) 2 n(n + 1)(2n + 1) 6 i=1   n X n(n + 1) 2 3 (∗ ∗ ∗) i = 2 (∗∗) i2 = i=1 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 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   6i 6 = . n n   n X 6i 6 f (xi )∆x = lim f n→∞ n n i=1  2 ! n X 6i 6 = 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 2 using (∗∗) 4 Z x − 1 dx using Riemann Sums and the Limit Definition of the Definite Integral. 2. Evaluate 0 2 Z x2 −5x dx using Riemann Sums and the Limit Definition of the Definite Integral. 3. Evaluate 0 Z 4 6 − 3x dx using Riemann Sums and the Limit Definition of the Definite Integral. 4. Evaluate 1 Turn in your own solutions for Problems 2, 3, and 4. 3