21b_ps_final_soln

# 21b_ps_final_soln - 21B Final Review Problems (Solutions)...

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Unformatted text preview: 21B Final Review Problems (Solutions) 1) x 3 + 6 x 2 + 2 x + 3 x 2 + 6 x + 1 ! dx First do long division to get: x dx + x + 3 x 2 + 6 x + 1 dx ! ! Note that the second integral can be done with u-substitution. Answer: x 2 2 + 1 2 ln x 2 + 6 x + 1 + C 2) x + 1 x 4 + x 2 dx ! = x + 1 x 2 ( x 2 + 1) dx ! Now use partial fractions: x + 1 x 2 ( x 2 + 1) = A x + B x 2 + Cx + D x 2 + 1 and solve for the constants. I got A = 1, B = 1, C = ! 1, D = ! 1 , so now we integrate 1 x + 1 x 2 + ! x ! 1 x 2 + 1 dx = ln x ! 1 x ! x + 1 x 2 + 1 dx " " (Factor out -1) To do the last integral, split it up: x + 1 x 2 + 1 dx = x x 2 + 1 dx + 1 x 2 + 1 dx = 1 2 ln( x 2 + 1) + tan ! 1 x " " " Final answer: ln x ! 1 x ! 1 2 ln( x 2 + 1) ! tan ! 1 x + C 3) ln(2 x ) x 2 dx ! Use Integration by Parts: u = ln(2 x ) ! du = 1 x dx , dV = 1 x 2 dx ! V = " 1 x ln(2 x ) x 2 dx ! = ! ln(2 x ) x ! ! 1 x " 1 x # dx = ! ln(2 x ) x ! 1 x + C 4) 1 x 3 + 1 dx ! Let u = x 3 + 1 ! x = ( u " 1) 3 ! dx = 3( u " 1) 2 du In terms of u , we now have 3( u ! 1) 2 du u " = 3 u 2 ! 2 u + 1 u " du = 3 u ! 2 + 1 u du = 3( u 2 2 ! 2 u + ln u ) + C " Now change back to x : = 3 ( x 3 + 1) 2 2 ! 2( x 3 + 1) + ln x 3 + 1 " # \$ % & ’ + C 5) x 3 4 ! x 2 dx " You can use trig substitution, but it is easier to do u-sub: u = 4 ! x 2 " du = ! 2 xdx Now “steal” the x you need to accommodate the du and write the remaining x 2 = 4 ! u = x 2 xdx u ! = " 1 2 4 " u u 1/2 du = " 1 2 4 u " 1/2 " u 1/2 du = " 1 2 8 u 1/2 " 2 3 u 3/2 # \$ % & ’ ( + C ! ! = ! 1 2 8(4 ! x 2 ) 1/2 ! 2 3 (4 ! x 2 ) 3/2 " # \$ % & ’ + C 6) dx x 4 x 2 ! 9 " Use trig. sub by letting x = 3sec ! " dx = 3sec ! tan ! d ! = 3sec ! tan ! d ! (3sec 4 ! ) 9sec 2 ! " 9 = 3 3 5 sec ! tan ! d ! sec 4 ! (sec 2 ! " 1) = # # 1 81 tan ! d ! sec 3 ! tan 2 ! # = 1 81 1 sec 3 ! d ! = 1 81 cos 3 ! d ! " " = 1 81 cos 2 ! cos ! d ! = 1 81 (1 " sin 2 ! )cos ! d ! = 1 81 (1 " u 2 ) du = 1 81 ( u " u 3 3 ) = 1 81 (sin ! " sin 3 ! 3 ) + C # # # Now draw a triangle that represents x = 3sec ! " sec ! = x 3 : x x 2 ! 9 ! 3 Write your answer in terms of x : 1 81 x 2 ! 9 x ! 1 3 x 2 ! 9 x " # \$ \$ % & ’ ’ 3 " # \$ \$ % & ’ ’ + C 7) Converge or diverge? a) x x 4 + 1 dx ! " = x x 4 + 1 dx 1 ! + x x 4 + 1 dx 1 " ! Now the first integral must converge since it is a proper definite integral, Use the Comparison Test for the 2 nd one: x x 4 + 1 dx < 1 x 3 dx 1 ! " 1 ! " , which converges, ( p > 1). So the whole thing converges. You can also evaluate the integral directly by letting u = x 2 ! du = 2 x dx , so the integral becomes 1 2 du u 2 + 1 ! " which becomes lim x !" 1 2 (tan # 1 u ) b = 1 2 \$ 2 # ( ) = !...
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## This note was uploaded on 05/26/2011 for the course MATH 16B taught by Professor Sarason during the Spring '06 term at Berkeley.

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21b_ps_final_soln - 21B Final Review Problems (Solutions)...

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