Calculus Solutions 14

Calculus Solutions 14 - Answers to Odd-Numbered Problems...

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Answers to Odd-Numbered Problems A-13 27x+1=u2,dx=2udu;$~=J[2-&]du=2~-21n(l+u)+C= 2,/2+1-21n(i+,/z+l)+c .-s + & by definition of derivative. At a double root Q'(a) = 0. 29 Note Q(o) = 0. Then = Section 7.5 Improper Integrals (page 309) 1-P 15 diverges for every p! 17 Less than $? 3 = + $PO ,q = tan-' XI; - -$]I" = + 2 19 Less than $,' ,& 21 Less than e-'dx = $, greater than -+ 23 Less than i,' e2dx + e $r e-('-')'dz = c2 + e $' e-~ldu = e2 + '- Jsr 3 $; + 1 less than - 25 1,' - + = 2 27 p! = p times (p - I)!; 1 = 1times 01 31 $; -2 Ldx = ifi 7~ + = G:- 33 w = 3pl~ - -- tmV; a, = 1000e--~~dt -10,000e-.~~]r = $10,000 = 29 u = x, dv = xe-"'dz : -x<]r $ = Jr ee--+ln2dx = C!II 00- 1 -In210 -m 35 37 $:I2 (see x - tan x)dx = [ln(secx + tan x) + ln(cos x)]:~' = [ln(l+ sin x)];l2 = In 2. The areas under sec x and tan x separately are infinite 39 Only p = 0 CHAPTER 8 APPLICATIONS OF THE INTEGRAL Section 8.1 Areas and Volumes by Slices (page 318) 1x2 - 3 = 1gives x = f2; ~!~[(1- 32 (x2 - 3)ldx = 7 3 3 = x = 9 gives y = f 3; $_S3[9 - y2]dy = 36 5 x4 - 2x2 = 2x2 gives x = f 2 (or x = 0); $!2[2x2
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