111_quiz5_spr08_key

111_quiz5_spr08_key - n = 3 to approximate ln4 = Z 4 1 dx x...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
APMA 111 Quiz 5 February 26, 2008 NAME: SECTION: #2 (11 am) #6 (10 am) On my honor as a student, I have neither given nor received aid on this quiz. SIGNATURE: 1. (a) [3 pts] Z e 2 dx Solution: Z e 2 dx = e 2 Z 1 dx = e 2 x + C (b) [3 pts] Z cos x cos 3 (sin x ) dx Solution: Set u = sin x , then du = cos xdx and Z cos x cos 3 (sin x ) dx = Z cos 3 (sin x ) cos xdx = Z cos 3 udu. Now use the identity cos 2 u = 1 - sin 2 u and set v = sin u : R cos 3 udu = R (1 - sin 2 u ) cos udu = R (1 - v 2 ) dv = v - 1 3 v 3 + C = sin u - 1 3 sin 3 u + C . Remembering that u = sin x , we get Z cos x cos 3 (sin x ) dx = sin(sin x ) - 1 3 sin 3 (sin x ) + C .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Question 1 continues . .. Quiz 5, Page 2 of 2 February 26, 2008 (c) [4 pts] Z 2 2 x 2 - 1 x dx Solution: Set x = sec θ , then dx = sec θ tan θ dθ and x 2 - 1 = sec 2 θ - 1 = tan 2 θ = tan θ . This implies that Z 2 2 x 2 - 1 x dx = Z π/ 3 π/ 4 (tan θ )( ± ± ± sec θ tan θ dθ ) ± ± ± sec θ = Z π/ 3 π/ 4 tan 2 θ dθ = Z π/ 3 π/ 4 (sec 2 θ - 1) = tan θ - θ ± ± ± ± ± π/ 3 π/ 4 = tan( π/ 3) - tan( π/ 4) - ( π/ 4 - π/ 3) = 3 - 1 - π/ 4 + π/ 3 = 12 3 - 12 + π 12 . 2. (a) [3 pts] Use the Trapezoidal Rule with
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n = 3 to approximate ln4 = Z 4 1 dx x . Solution: Since Δ x = ( b-a ) /n = (4-1) / 3 = 1 and f ( x ) = 1 /x , we have the following: i : 0 1 2 3 x i : 1 2 3 4 f ( x i ) : 1 1 / 2 1 / 3 1 / 4 w i : 1 2 2 1 (these are the coefficients or weights ) p i : 1 1 2 / 3 1 / 4 (note that p i = w i · f ( x i )) This implies that T 3 = 4-1 2 · 3 [ f ( x ) + 2 f ( x 1 ) + 2 f ( x 2 ) + f ( x 3 )] = 1 2 ( 1 + 1 + 2 3 + 1 4 ) = 12+12+8+3 2 · 12 = 35 / 24 . (b) [2 pts] Is this an underestimate or an overestimate? Solution: The trapezoids are above the curve, so this is an overestimate....
View Full Document

This note was uploaded on 09/21/2009 for the course MATH 1234 taught by Professor Egcdd during the Spring '09 term at Aarhus Universitet.

Page1 / 2

111_quiz5_spr08_key - n = 3 to approximate ln4 = Z 4 1 dx x...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online