APPM 1350 Review Solutions 3

# 6 i z z 2 i1 8 given that n x i1 using left

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Unformatted text preview: that: n X i=1 using left endpts n(n + 1) i= , 2 n X i=1 using right endpts n(n + 1)(2n + 1) i= , 6 2 n X and 3 i= i=1 n(n + 1) 2 2 . Use the limit deﬁnition of the integral, with right endpoints and Z regular partition, to evaluate the a Z2 1 2 following integrals, show all work: (a) (2 x ) dx (b) x3 + 3x dx 0 0 Solution: (a) Note that Z 2 x = (2 2 (2 x ) dx |{z} lim = n!1 0 x⇤ = xi i 0)/n = 2/n and xi = 0 + i x = 2i/n and so, n X f ( xi ) x = lim n!1 i=1 = lim n!1 = lim n!1 = (b) Here, Z 1 0 x = (1 lim n!1 n X✓ 4i 2 n2 2 i=1 &quot; n 4X 1 n i=1 4 n n 4 ◆ 2 n n 8X2 i n3 i=1 # 8 n(n + 1)(2n + 1) n3 6 3 + (lower order terms) 16n =4 6n 3 16 = 8/6 = 4/3 6 0)/n = 1/n and xi = 0 + i x = i/n and so, 3 (x + 3x) dx |{z} lim = n!1 x⇤ = xi i n X f ( xi ) x = i=1 = lim n!1 lim n!1 n X ✓ i3 i=1 &quot; 3i + n3 n ◆ 1 n n n 1X3 3X i+ 2 i n4 n i=1 n2 ( n i=1 # 1)2 1 + 3 n(n + 1) +2 n4 4 n 2 4 n + (lower order terms) 3n2 + 3n 13 = lim + = + = 7/4 4 2 n!1 4n 2n 42 = 9. (a) Write the sum in expanded form 10 X k=6 k x, lim n!1 (b) Evaluat...
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## This note was uploaded on 02/04/2014 for the course APPM 1350 taught by Professor Segur,harv during the Fall '07 term at Colorado.

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