Calc 1 WA 5 - WA 5 Calculus I Chapter 5 Section 5.1 dr = d...

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Unformatted text preview: WA 5 Calculus I Chapter 5 Section 5.1 dr = d 6) r = d = d 1 dx x2 x -2 dx 10) x -1 +c -1 1 +c -x x( x 2 + 3)dx 12) x 3 + 3 dx = x 4 3x 2 + +c 4 2 (t 2 - sin t )dt t3 + cos t + c 3 32) check ... = d'= 3t 2 + ( - sin t ) = t 2 - sin t 3 sec y (tan y - sec y )dy (tan y )(sec y ) - sec 2 y dy 38) = sec y - tan y + c check ... d ' = (tan y )(sec y ) - sec 2 y = sec y (tan y - sec y ) dy = 2 x - 2 (3, 2) dx 2x2 y= - 2x + c 2 54) 2 (32 ) 2= - 2(3) + c - > c = -1 2 y = x2 - 2x -1 dy 3 = x > 0 (e,3) dx x 1 dy 3 x dx 56) y = 3ln x + c 3 = ln e + c - > c = 3 y = 3ln e + 3 Should be 3lnx+C f ''( x) = sin x f '( x ) = sin x dx = - cos x + c1 f '(0) = - cos 0 + c1 = 1 - > c1 = 2 70) f ( x ) = - cos x + 2 = sin x + 2 x + c2 f (0) = 6 = sin(0) + 2(0) + c2 = 6 6 = 0 + 0 + c2 - > c2 = 6 dp = k t where 100 = k 1 so100 = K dt 10 k t dt 0 10 k (t 1/2 ) dt 0 10 100 t 1/2 dt 0 74) 2t 3/2 ) + c1 3 2(0)3/2 500 = 100( ) + c1 3 c1 = 500 P = 100( @ day 7 P = 100( 2(7)3/2 ) + 500 3 P = 1, 734.68 a (t ) = -32 ft / sec 2 v (t ) = -32t + v0 s f = (1/ 2)at 2 + v0 (t ) + s0 ft 2 ft )t + (8 )(t ) + 64 ft 2 sec sec ft ft 80) 0 = 16( 2 )(t 2 ) + 8 (t ) + 64 ft sec sec t = 2.2655 sec ft v (t ) = -32( 2 )(2.2655sec) + 8 sec ft = -64.496 sec 0 = (1/ 2)( -32 Section 5.2 6 k =3 k (k - 2) = 3(3 - 2) + 4(4 - 2) + 5(5 - 2) + 6(6 - 2) 2) = 3 + 8 + 15 + 24 = 50 1 2 4 2 1 .... 1 - ( 4 ) + - ( 4 ) 10) 1 V - ( 4 ) x =1 4 x 2 15 I =1 ( 2i - 3) = 15 i =1 2 - = i 3 16) i =1 15 15(15 + 1) 2( ) - (15(3)) = 2 2(120) - 45 = 240 - 45 = 195 Section 5.3 3 3x 2 dx 1 6) Vx = (3 - 1) = 2 n n Vx = 0 as x - >8 (8 - x)dx 0 h = 8, b = 8 28) triangle 1 A = (bh) 2 1 A = (8)(8) = 32 2 8 A = (8 - x) dx = 32 0 1 2 1 x ) - (8 x - x 2 ) 2 2 1 2 1 = (8)(8 - ( 8 ) ) - (8)(0) - ( )(02 ) 2 2 = 64 - 32 - 0 = 32 = (8 x - 44) a) -1 f f f ( x)dx = ( x)dx + ( ( x)dx) -1 -1 0 1 0 = 0 + (-5) = -5 b) -1 f f f ( x)dx = ( x)dx + (- ( x)dx) -1 1 0 1 0 = 5 - ( -5) = 10 1 1 3 f( c) f ( x)dx = 3 x)dx = 3(0) = 0 3 f( d) f ( x)dx = 3 x) dx = 3(5) = 15 0 0 -1 1 -1 1 Section 5.4 3 5 x 2 (3 x + 5 x - 4)dx = + x - 4x 2 1 1 2 3 3 10) = (33 + 5(3) - 4(3)) - ((13 ) + 5(1) - 4(1)) 2 2 1 43 = 22 - ( ) = 2 2 Answer here should be 38 2 1 1 u (u - 2 )du = u - 1u -2 du = - u 2 -2 u -2 (( (-1) 2 1 (-2) 2 1 - ( )) - ( - ( )) 2 -1 2 -2 = -1 -1 -1 2 14) Answer here should be -2 8 1 - 1 2 dx = 2 2 dx = 2 x 2 x 2 x 1 1 1 8 8 18) = 2(2( 8)) - 2(2( 1)) = 2((2 8) - (2 1)) = 2 2( 8 - 1) /4 1 - sin 2 cos2 d = 0 /4 /4 1 - sin 2 - sin 2 = 1d 1 0 28) given : cos 2 + sin 2 = 1, then 1 - sin 2 = cos 2 [ ]0 ( ) - (0) = ( ) 4 4 ( x + 1) x 1 1 1dx x dx = + x dx = + dx x x 1 1 1 1 ln 1 32) = [ x ] 1 + x = [ 5] - [ 1] + [ ln 5] - [ ln1] 5 5 5 5 5 5 = 5 - 1 + ln 5 - 0 = 4 - ln 5 2e 1 1 (cos cos x - x )dx = xdx - dx x e e e 2e 2e 2e 38) = sin x 2e - x ln e [ ]e = (sin 2e - sin e) - ((ln 2e ) - (1)) y = 1- x 4 -1 1 (1 1dx x - x )dx = - dx 4 4 -1 -1 1 1 1 5 x 1 40) = [ x ] -1 - 5 -1 15 -15 1 -1 2 = (1 - 1) - ( - ) = 0- - = 5 5 5 5 5 Answer here should be 8/5 Section 5.5 x (4 x 2 + 3) 2 dx u = 4 x2 + 3 V 14) du = 8 xdx 1 2 2 8 V x (4 x + 3) dx 8 1 2 = V du u 8 1 u3 =( ) +c 8 3 = 1 (4 x 2 + 3)3 ( )+c 8 3 (4 x 2 + 3)3 1 dx ) ( + c x 2 + 3) 2 x = (4 8 3 t + 2t 2 2 2 (t t dt = + 2t )t dt = V1/2 + 2t 3/2 dt t = 2 4 t ( t + t2) + c 3 5 4 2 dx t ( t + t 2 ) + c 30) 5 3 2 4 = t 3/2 + t 5/2 + c 3 5 1/2 3/2 = t + 2t = (t + 2t 2 )t -1/2 = t + 2t 2 t 2 x 2x sin x dx = 2 sin x dx 2 1 u = x2 du = 2 xdx = 52) 1 sin udu 2V 1 = - cos u + c 2 1 = (- cos x 2 ) + c 2 1 1 dx ( - cos x 2 ) + c sin x 2 2 x = 2 2 = x sin x 2 (x + 1)e x2 + 2 x u = x2 + 2x 56) du = 2 x + 2 dx 1 x2 +2 x 2( x + 1)e dx 2 1 u e du 2V 1 = eu + c 2 1 2 = e x +2 x + c 2 2 1 2 1 2 dx e x + 2 x + c (e x + 2 x )( 2 x + 2 ) = (e x + 2 x )( x + 1) = 2 2 sin x -3 (sin 3 x dx = x)(cos x)dx cos - cos 2 x 1 ( - cos 2 x) =( )( )+c = +c 2 -2 sin 2 x (-4sin 2 x) -1 cot 2 x - cot 2 x )+c = +c 62) = ( )( -4 1 -4 cot 2 x cos 2 x - cos 2 x)(sin -2 x) - - ( dx + c = + c = + c = 2 - -4 -4 4sin x ( -2 sin x)( -2 cos -3 x) sin x = cos3 x -4 x 2 x + 3dx du dx u -3 du x= dx = 2 2 u - 3 1/2 du 1 1/2 ( (u 3/2 2 )(u ) 2 = 4 - 3u )du 88) 1 2 1 = ( u 5/2 - u 3/2 ) + c 4 5 2 1 1 = u 5/2 - u 3/2 + c 10 8 1 1 3/2 = (2 x + 3)5/2 - ( 2 x + 3) + c 10 8 u = 2x + 3 2= (2 x + 4 dx = x + 1)( x + 4) u = x+4 x =u-4 1/2 -1/2 2x +1 -1/2 dx dx = du -1/2 92) (2( u - 4) + 1) (u 2 Vu - 7u du )du = u - 7)u -1/2 du (2 4 3/2 u - 14u1/2 + c 3 = 4 + u u - 14 C 3 4 x + 4( ( x + 4) - 14) + c 3 e1- x eu eu e1- x dx = du = = 1 1 2 2 2 2 102) 1 1 u = 1- x du = du 2 = e1- 2 - e1-1 = e -1 - 1 2 x3 4 + x dx = 0 1 2 x 3 4 + x 2 dx 2 0 1 3 udu 2V 0 1 1/3 u du 2V 0 2 2 2 2 u = 4 + x2 du = 2 xdx = = 1 3 = u 4/3 106) 2 4 2 3 = + x 2 ) 4/3 (4 0 8 3 3 = + 22 ) 4/3 + 02 ) 4/3 (4 - (4 8 A 8 3 3 = (8) 4/3 - (4) 4/3 8 8 3 = (16 - 3 44 ) 8 /2 /3 (x xdx cos + cos x)dx = + xdx /3 /3 2 /2 /2 x /2 + [ sin x ] /3 2 /3 110) = ( 1 )(( ) 2 - ( ) 2 ) + (sin ) - (sin ) 2 2 3 2 3 2 2 1 = (sin ) - (sin ) + - 2 2 3 4 9 /2 2 2 = ( - ) + (sin ) - (sin ) 8 18 2 3 Section 5.7 1 1 1 3( + 2 dx = 3 3x + 2 )dx 3x 1 1 u = 3 x + 2 = V du 3 u 6) 1 du = 3dx = ( ln u ) + c 3 1 = (ln 3 x + 2 ) + c 3 x2 1 -3 x 2 1 1 1 dx = - 3 dx = - du = - (ln u ) + c - x3 3 3 3- x 3 u 3 8) 1 u = 3 - x3 = - (ln 3 - x 3 ) + c 3 du = -3 x 1 1 1 x 1 dx ln( x3 ) dx = ln( x3 ) = xx3 dx 1 x x ln x 20) u = ln x du = 1 u1 3 dx = V 3 dx = ln u + c x ln u 3 = ln ln u + c tan - 5 d = 5 d = - cos 5 d cos u' u = cos 5 = - V d = - ln u + c 30) u du = - sin 5 d 1 1 sin 5 sin 5 = -ln cos 5 + c 1 1 x -1 2 1 1 1 u' 1- + 1dx = x + 1 = 2- x + 1 dx = 2- u dx x 2 2 0 0 0 0 u = x +1 54) du = dx 1 = 2 - ln u 2 0 1 = 2 - ln x + 1 2 0 1 1 1 1 2( - ln 1 + 1) - 2( - ln 0 + 1) 2 2 1 - 2 ln 2 - 1 = -2 ln 2 ...
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