Question 1

Solve the initial value problem for y as a function of x.
- xdydx=x2−9,x≥3,y(3) =0x \frac { d y } { d x } = \sqrt { x ^ { 2 } - 9 } , x \geq 3 , y ( 3 )  = 0xdxdy​=x2−9​,x≥3,y(3) =0

Accepted Answer

A)    y=3ln⁡x3y = 3 \ln \frac { x } { 3 }y=3ln3x​ 
B)    y=x2−93−x+3y = \frac { \sqrt { x ^ { 2 } - 9 } } { 3 } - x + 3y=3x2−9c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​−x+3 
C)    y=x2−93sec⁡−1(x/3) y = \frac { \sqrt { x ^ { 2 } - 9 } } { 3 \sec ^ { - 1 } ( x / 3 )  }y=3sec−1(x/3) x2−9c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​ 
D)    y=3[x2−93−sec⁡−1(x3) ]y = 3 \left[ \frac { \sqrt { \mathrm { x } ^ { 2 } - 9 } } { 3 } - \sec ^ { - 1 } \left( \frac { \mathrm { x } } { 3 } \right)  \right]y=3[3x2−9c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​−sec−1(3x​) ]

Question 2

Evaluate the improper integral.
- ∫025dx∣x−16∣\int _ { 0 } ^ { 25 } \frac { d x } { \sqrt { | x - 16 | } }∫025​∣x−16∣​dx​

Accepted Answer

A)   6
B)   7
C)   14
D)   -2

Question 3

Provide an appropriate response.
-  The "Simpson" sum is based upon the area under a ?‾. 	ext { The "Simpson" sum is based upon the area under a } \underline { ? } 	ext {. } The "Simpson" sum is based upon the area under a ?​.

Accepted Answer

A)   rectangle
B)   trapezoid
C)   triangle
D)   parabola

Question 4

Use reduction formulas to evaluate the integral.
- ∫8cos⁡32xdx\int 8 \cos ^ { 3 } 2 x d x∫8cos32xdx

Accepted Answer

A)    4sin⁡2x−43cos⁡32x+C4 \sin 2 x - \frac { 4 } { 3 } \cos ^ { 3 } 2 x + C4sin2x−34​cos32x+C 
B)    4sin⁡2x+43sin⁡32x+C4 \sin 2 x + \frac { 4 } { 3 } \sin ^ { 3 } 2 x + C4sin2x+34​sin32x+C 
C)    8sin⁡2x−83sin⁡32x+C8 \sin 2 x - \frac { 8 } { 3 } \sin ^ { 3 } 2 x + C8sin2x−38​sin32x+C 
D)    4sin⁡2x−43sin⁡32x+C4 \sin 2 x - \frac { 4 } { 3 } \sin ^ { 3 } 2 x + C4sin2x−34​sin32x+C

Question 5

Solve the problem.
-Suppose a brewery has a filling machine that fills 12 ounce bottles of beer. It is known that the amount of beer poured by this filling machine follows a normal distribution with a mean of 12.14 onces and a standard
Deviation of 0.04 ounce. Find the probability that the bottle contains more than 12.14 ounces of beer.

Accepted Answer

A)   0.4
B)   1
C)   0
D)   0.5

Question 6

Evaluate the integral.
- ∫1xln⁡x6dx\int \frac { 1 } { x \ln x ^ { 6 } } d x∫xlnx61​dx

Accepted Answer

A)    ln⁡(ln⁡x6) +C\ln \left( \ln x ^ { 6 } \right)  + Cln(lnx6) +C 
B)    ln⁡x6+C\ln x ^ { 6 } + Clnx6+C 
C)    16ln⁡x6+C\frac { 1 } { 6 } \ln x ^ { 6 } + C61​lnx6+C 
D)    16ln⁡(ln⁡x6) +C\frac { 1 } { 6 } \ln \left( \ln x ^ { 6 } \right)  + C61​ln(lnx6) +C

Question 7

Evaluate the integral by making a substitution (possibly trigonometric)  and then applying a reduction formula.
- ∫dx(25−x2) 2\int \frac { d x } { \left( 25 - x ^ { 2 } \right)  ^ { 2 } }∫(25−x2) 2dx​

Accepted Answer

A)    110ln⁡∣x+5x−5∣+C\frac { 1 } { 10 } \ln \left| \frac { x + 5 } { x - 5 } \right| + C101​lnc10,0,16.667,-5,20,-15 v-585 v0 v-585 c-2.667,-10,-9.667,-15,-21,-15c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v0 v585 h43z'/>​x−5x+5​c10,0,16.667,-5,20,-15 v-585 v0 v-585 c-2.667,-10,-9.667,-15,-21,-15c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v0 v585 h43z'/>​+C 
B)    x50(25−x2) +C\frac { x } { 50 \left( 25 - x ^ { 2 } \right)  } + C50(25−x2) x​+C 
C)    150(x25−x2+110ln⁡∣x+5x−5∣) +C\frac { 1 } { 50 } \left( \frac { x } { 25 - x ^ { 2 } } + \frac { 1 } { 10 } \ln \left| \frac { x + 5 } { x - 5 } \right| \right)  + C501​(25−x2x​+101​lnc10,0,16.667,-5,20,-15 v-585 v0 v-585 c-2.667,-10,-9.667,-15,-21,-15c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v0 v585 h43z'/>​x−5x+5​c10,0,16.667,-5,20,-15 v-585 v0 v-585 c-2.667,-10,-9.667,-15,-21,-15c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v0 v585 h43z'/>​) +C 
D)    150(x25−x2−110ln⁡∣x+5x−5∣) +C\frac { 1 } { 50 } \left( \frac { x } { 25 - x ^ { 2 } } - \frac { 1 } { 10 } \ln \left| \frac { x + 5 } { x - 5 } \right| \right)  + C501​(25−x2x​−101​lnc10,0,16.667,-5,20,-15 v-585 v0 v-585 c-2.667,-10,-9.667,-15,-21,-15c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v0 v585 h43z'/>​x−5x+5​c10,0,16.667,-5,20,-15 v-585 v0 v-585 c-2.667,-10,-9.667,-15,-21,-15c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v0 v585 h43z'/>​) +C

Question 8

Evaluate the integral.
-Use the formula  ∫f−1(x) dx=xf−1(x) −∫x(ddxf−1(x) ) dx\int f ^ { - 1 } ( x )  d x = x f ^ { - 1 } ( x )  - \int x \left( \frac { d } { d x } f ^ { - 1 } ( x )  \right)  d x∫f−1(x) dx=xf−1(x) −∫x(dxd​f−1(x) ) dx  to evaluate the integral.  ∫cot⁡−1xdx\int \cot ^ { - 1 } x d x∫cot−1xdx

Accepted Answer

A)    xcot⁡−1x+ln⁡x+Cx \cot ^ { - 1 } x + \ln x + Cxcot−1x+lnx+C 
B)    xcot⁡−1x+x+Cx \cot ^ { - 1 } x + x + Cxcot−1x+x+C 
C)    xcot⁡−1x+ln⁡x2+1+Cx \cot ^ { - 1 } x + \ln \sqrt { x ^ { 2 } + 1 } + Cxcot−1x+lnx2+1c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​+C 
D)    xcot⁡−1x−ln⁡x2+1+Cx \cot ^ { - 1 } x - \ln \sqrt { x ^ { 2 } + 1 } + Cxcot−1x−lnx2+1c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​+C

Question 9

Solve the problem by integration.
-The force  F\mathrm { F }F  (in N)  applied by a stamping machine in making a certain computer part is  F=3xx2+11x+12\mathrm { F } = \frac { 3 \mathrm { x } } { \mathrm { x } ^ { 2 } + 11 \mathrm { x } + 12 }F=x2+11x+123x​ , where  x\mathrm { x }x  is the distance (in  cm\mathrm { cm }cm  )  through which the force acts. Find the work done by the force from  x=0\mathrm { x } = 0x=0  to  x=0.6 cm\mathrm { x } = 0.6 \mathrm {~cm}x=0.6 cm .

Accepted Answer

A)   0.0832 N· cm
B)   8.0928 N· cm
C)   0.0166 N· cm
D)   0.0055 N· cm

Question 10

Solve the problem.
-Find an upper bound for  ∣ET∣\left| \mathrm { E } _ { \mathrm { T } } \right|∣ET​∣  in estimating  ∫15(3x2+8) dx\int _ { 1 } ^ { 5 } \left( 3 \mathrm { x } ^ { 2 } + 8 \right)  \mathrm { dx }∫15​(3x2+8) dx  with  n=6\mathrm { n } = 6n=6  steps.

Accepted Answer

A)    12572\frac { 125 } { 72 }72125​ 
B)    89\frac { 8 } { 9 }98​ 
C)    49\frac { 4 } { 9 }94​ 
D)    427\frac { 4 } { 27 }274​

Question 11

Use various trigonometric identities to simplify the expression then integrate.
- ∫cos⁡7θsin⁡2θdθ\int \cos ^ { 7 } 	heta \sin 2 	heta d 	heta∫cos7θsin2θdθ

Accepted Answer

A)    14cos⁡7θ+C\frac { 1 } { 4 } \cos ^ { 7 } 	heta + C41​cos7θ+C 
B)    −15cos⁡10θ+C- \frac { 1 } { 5 } \cos ^ { 10 } 	heta + C−51​cos10θ+C 
C)    19cos⁡8θ+C\frac { 1 } { 9 } \cos ^ { 8 } 	heta + C91​cos8θ+C 
D)    −29cos⁡9θ+C- \frac { 2 } { 9 } \cos ^ { 9 } 	heta + C−92​cos9θ+C

Question 12

Solve the initial value problem for y as a function of x.
-  2 , y ( 4 )  = 0">x2−4dydx=x,x>2,y(4) =0\sqrt { x ^ { 2 } - 4 } \frac { d y } { d x } = x , x > 2 , y ( 4 )  = 0x2−4​dxdy​=x,x>2,y(4) =0

Accepted Answer

A)    y=x2−4−23y = \sqrt { x ^ { 2 } - 4 } - 2 \sqrt { 3 }y=x2−4c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​−23c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​ 
B)    y=x2−4y = \sqrt { x ^ { 2 } - 4 }y=x2−4c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​ 
C)    y=x2−4xy = \frac { \sqrt { x ^ { 2 } - 4 } } { x }y=xx2−4c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​ 
D)    y=x2−4x−32y = \frac { \sqrt { x ^ { 2 } - 4 } } { x } - \frac { \sqrt { 3 } } { 2 }y=xx2−4c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​−23c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​

Question 13

Use reduction formulas to evaluate the integral.
- ∫sin⁡23xcos⁡23xdx\int \sin ^ { 2 } 3 x \cos ^ { 2 } 3 x d x∫sin23xcos23xdx

Accepted Answer

A)    −13sin⁡3xcos⁡33x+x8+13sin⁡6x+C- \frac { 1 } { 3 } \sin 3 x \cos ^ { 3 } 3 x + \frac { x } { 8 } + \frac { 1 } { 3 } \sin 6 x + C−31​sin3xcos33x+8x​+31​sin6x+C 
B)    −112sin⁡3xcos⁡23x+x8+148sin⁡6x+C- \frac { 1 } { 12 } \sin 3 x \cos ^ { 2 } 3 x + \frac { x } { 8 } + \frac { 1 } { 48 } \sin 6 x + C−121​sin3xcos23x+8x​+481​sin6x+C 
C)    −112sin⁡3xcos⁡33x+x8+148sin⁡6x+C- \frac { 1 } { 12 } \sin 3 x \cos ^ { 3 } 3 x + \frac { x } { 8 } + \frac { 1 } { 48 } \sin 6 x + C−121​sin3xcos33x+8x​+481​sin6x+C 
D)    −112sin⁡3xcos⁡33x+x8+148cos⁡6x+C- \frac { 1 } { 12 } \sin 3 x \cos ^ { 3 } 3 x + \frac { x } { 8 } + \frac { 1 } { 48 } \cos 6 x + C−121​sin3xcos33x+8x​+481​cos6x+C

Question 14

Determine whether the improper integral converges or diverges.
- ∫−∞∞dx5x6+1\int _ { - \infty } ^ { \infty } \frac { d x } { \sqrt { 5 x ^ { 6 } + 1 } }∫−∞∞​5x6+1​dx​

Accepted Answer

A)   Diverges
B)   Converges

Question 15

Determine whether the improper integral converges or diverges.
- ∫1∞dxx2/5+2\int _ { 1 } ^ { \infty } \frac { d x } { x ^ { 2 / 5 } + 2 }∫1∞​x2/5+2dx​

Accepted Answer

A)   Converges
B)   Diverges

Question 16

Use Simpson's Rule with n = 4 steps to estimate the integral.
- ∫−10sin⁡πtdt\int _ { - 1 } ^ { 0 } \sin \pi \mathrm { tdt }∫−10​sinπtdt

Accepted Answer

A)    −1+24- \frac { 1 + \sqrt { 2 } } { 4 }−41+2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​ 
B)    −1−22- 1 - 2 \sqrt { 2 }−1−22c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​ 
C)    −1+226- \frac { 1 + 2 \sqrt { 2 } } { 6 }−61+22c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​ 
D)    −2+26- \frac { 2 + \sqrt { 2 } } { 6 }−62+2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​

Question 17

Integrate the function.
-  \frac { 11 } { 5 }">∫dx25x2−121,x>115\int \frac { d x } { \sqrt { 25 x ^ { 2 } - 121 } } , x > \frac { 11 } { 5 }∫25x2−121​dx​,x>511​

Accepted Answer

A)    111ln⁡∣115x+25x2−1215∣+C\frac { 1 } { 11 } \ln \left| \frac { 11 } { 5 } x + \frac { \sqrt { 25 x ^ { 2 } - 121 } } { 5 } \right| + C111​lnc10,0,16.667,-5,20,-15 v-585 v-600 v-585 c-2.667,-10,-9.667,-15,-21,-15c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v600 v585 h43z'/>​511​x+525x2−121c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​c10,0,16.667,-5,20,-15 v-585 v-600 v-585 c-2.667,-10,-9.667,-15,-21,-15c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v600 v585 h43z'/>​+C 
B)    15ln⁡∣sec⁡−1(115x) +25x2−12111∣+C\frac { 1 } { 5 } \ln \left| \sec ^ { - 1 } \left( \frac { 11 } { 5 } x \right)  + \frac { \sqrt { 25 x ^ { 2 } - 121 } } { 11 } \right| + C51​lnc10,0,16.667,-5,20,-15 v-585 v-600 v-585 c-2.667,-10,-9.667,-15,-21,-15c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v600 v585 h43z'/>​sec−1(511​x) +1125x2−121c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​c10,0,16.667,-5,20,-15 v-585 v-600 v-585 c-2.667,-10,-9.667,-15,-21,-15c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v600 v585 h43z'/>​+C 
C)    15ln⁡∣511x+25x2−12111∣+C\frac { 1 } { 5 } \ln \left| \frac { 5 } { 11 } x + \frac { \sqrt { 25 x ^ { 2 } - 121 } } { 11 } \right| + C51​lnc10,0,16.667,-5,20,-15 v-585 v-600 v-585 c-2.667,-10,-9.667,-15,-21,-15c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v600 v585 h43z'/>​115​x+1125x2−121c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​c10,0,16.667,-5,20,-15 v-585 v-600 v-585 c-2.667,-10,-9.667,-15,-21,-15c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v600 v585 h43z'/>​+C 
D)    15ln⁡∣52x+1125x2−121∣+C\frac { 1 } { 5 } \ln \left| \frac { 5 } { 2 } x + \frac { 11 } { \sqrt { 25 x ^ { 2 } - 121 } } \right| + C51​lnc10,0,16.667,-5,20,-15 v-585 v-600 v-585 c-2.667,-10,-9.667,-15,-21,-15c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v600 v585 h43z'/>​25​x+25x2−121c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​11​c10,0,16.667,-5,20,-15 v-585 v-600 v-585 c-2.667,-10,-9.667,-15,-21,-15c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v600 v585 h43z'/>​+C

Question 18

Find the value of the constant k so that the given function in a probability density function for a random variable over the
specified interval.
- f(x) =kxf ( x )  = k ^ { x }f(x) =kx  over  [0,20][ 0,20 ][0,20]

Accepted Answer

A)    1e20−1\frac { 1 } { \mathrm { e } ^ { 20 } - 1 }e20−11​ 
B)    1e20+1\frac { 1 } { \mathrm { e } ^ { 20 } + 1 }e20+11​ 
C)    e20−1\mathrm { e } ^ { 20 } - 1e20−1 
D)    2e20\frac { 2 } { \mathrm { e } ^ { 20 } }e202​

Question 19

Provide an appropriate response.
-(a) Find the values of $p$ for which $\int _ { 0 } ^ { 1 } \frac { 1 } { x p } d x$ converges.
(b) Find the values of $p$ for which $\int _ { 0 } ^ { 1 } \frac { 1 } { x _ { p } } d x$ diverges.

Accepted Answer

(a) The integral con...

Question 20

Expand the quotient by partial fractions.
- x+4x2+2x+1\frac { x + 4 } { x ^ { 2 } + 2 x + 1 }x2+2x+1x+4​

Accepted Answer

A)    3x+1+1(x+1) 2\frac { 3 } { x + 1 } + \frac { 1 } { ( x + 1 )  ^ { 2 } }x+13​+(x+1) 21​ 
B)    1x+1+4x+4\frac { 1 } { x + 1 } + \frac { 4 } { x + 4 }x+11​+x+44​ 
C)    1x+1+3(x+1) 2\frac { 1 } { x + 1 } + \frac { 3 } { ( x + 1 )  ^ { 2 } }x+11​+(x+1) 23​ 
D)    1x+1+−3(x+1) 2\frac { 1 } { x + 1 } + \frac { - 3 } { ( x + 1 )  ^ { 2 } }x+11​+(x+1) 2−3​

Exam 9: Techniques of Integration

Evaluate the improper integral. - $\int _ { 0 } ^ { 25 } \frac { d x } { \sqrt { | x - 16 | } }$

Correct Answer
verified

Find the value of the constant k so that the given function in a probability density function for a random variable over the specified interval. - $f ( x ) = k ^ { x }$ over $[ 0,20 ]$

Correct Answer
verified

Solve the initial value problem for y as a function of x. - $\sqrt { x ^ { 2 } - 4 } \frac { d y } { d x } = x , x > 2 , y ( 4 ) = 0$

Correct Answer
verified

Correct Answer
verified

Integrate the function. - $\int \frac { d x } { \sqrt { 25 x ^ { 2 } - 121 } } , x > \frac { 11 } { 5 }$

Correct Answer
verified

Determine whether the improper integral converges or diverges. - $\int _ { - \infty } ^ { \infty } \frac { d x } { \sqrt { 5 x ^ { 6 } + 1 } }$

Correct Answer
verified

Evaluate the integral by making a substitution (possibly trigonometric) and then applying a reduction formula. - $\int \frac { d x } { \left( 25 - x ^ { 2 } \right) ^ { 2 } }$

Correct Answer
verified

Use Simpson's Rule with n = 4 steps to estimate the integral. - $\int _ { - 1 } ^ { 0 } \sin \pi \mathrm { tdt }$

Correct Answer
verified

Determine whether the improper integral converges or diverges. - $\int _ { 1 } ^ { \infty } \frac { d x } { x ^ { 2 / 5 } + 2 }$

Correct Answer
verified

Use reduction formulas to evaluate the integral. - $\int \sin ^ { 2 } 3 x \cos ^ { 2 } 3 x d x$

Correct Answer
verified

Solve the problem. -Find an upper bound for $\left| \mathrm { E } _ { \mathrm { T } } \right|$ in estimating $\int _ { 1 } ^ { 5 } \left( 3 \mathrm { x } ^ { 2 } + 8 \right) \mathrm { dx }$ with $\mathrm { n } = 6$ steps.

Correct Answer
verified

Expand the quotient by partial fractions. - $\frac { x + 4 } { x ^ { 2 } + 2 x + 1 }$

Correct Answer
verified

Correct Answer
verified

Solve the initial value problem for y as a function of x. - $x \frac { d y } { d x } = \sqrt { x ^ { 2 } - 9 } , x \geq 3 , y ( 3 ) = 0$

Correct Answer
verified

Use reduction formulas to evaluate the integral. - $\int 8 \cos ^ { 3 } 2 x d x$

Correct Answer
verified

Evaluate the integral. - $\int \frac { 1 } { x \ln x ^ { 6 } } d x$

Correct Answer
verified

Evaluate the integral. -Use the formula $\int f ^ { - 1 } ( x ) d x = x f ^ { - 1 } ( x ) - \int x \left( \frac { d } { d x } f ^ { - 1 } ( x ) \right) d x$ to evaluate the integral. $\int \cot ^ { - 1 } x d x$

Correct Answer
verified

Provide an appropriate response. -(a) Find the values of $p$ for which $\int _ { 0 } ^ { 1 } \frac { 1 } { x p } d x$ converges. (b) Find the values of $p$ for which $\int _ { 0 } ^ { 1 } \frac { 1 } { x _ { p } } d x$ diverges.

Correct Answer
verified
(a) The integral con...
View Answer

Provide an appropriate response. - $\text { The "Simpson" sum is based upon the area under a } \underline { ? } \text {. }$

Correct Answer
verified

Use various trigonometric identities to simplify the expression then integrate. - $\int \cos ^ { 7 } \theta \sin 2 \theta d \theta$

Correct Answer
verified

Exam 9: Techniques of Integration

Evaluate the improper integral. - ∫025dx∣x−16∣\int _ { 0 } ^ { 25 } \frac { d x } { \sqrt { | x - 16 | } }∫025​∣x−16∣​dx​

Correct AnswerverifiedShow Answer

Find the value of the constant k so that the given function in a probability density function for a random variable over the specified interval. - f(x) =kxf ( x ) = k ^ { x }f(x) =kx over [0,20][ 0,20 ][0,20]

Correct AnswerverifiedShow Answer

Solve the initial value problem for y as a function of x. - x2−4dydx=x,x>2,y(4) =0\sqrt { x ^ { 2 } - 4 } \frac { d y } { d x } = x , x > 2 , y ( 4 ) = 0x2−4​dxdy​=x,x>2,y(4) =0

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Integrate the function. - ∫dx25x2−121,x>115\int \frac { d x } { \sqrt { 25 x ^ { 2 } - 121 } } , x > \frac { 11 } { 5 }∫25x2−121​dx​,x>511​

Correct AnswerverifiedShow Answer

Determine whether the improper integral converges or diverges. - ∫−∞∞dx5x6+1\int _ { - \infty } ^ { \infty } \frac { d x } { \sqrt { 5 x ^ { 6 } + 1 } }∫−∞∞​5x6+1​dx​

Correct AnswerverifiedShow Answer

Evaluate the integral by making a substitution (possibly trigonometric) and then applying a reduction formula. - ∫dx(25−x2) 2\int \frac { d x } { \left( 25 - x ^ { 2 } \right) ^ { 2 } }∫(25−x2) 2dx​

Correct AnswerverifiedShow Answer

Use Simpson's Rule with n = 4 steps to estimate the integral. - ∫−10sin⁡πtdt\int _ { - 1 } ^ { 0 } \sin \pi \mathrm { tdt }∫−10​sinπtdt

Correct AnswerverifiedShow Answer

Determine whether the improper integral converges or diverges. - ∫1∞dxx2/5+2\int _ { 1 } ^ { \infty } \frac { d x } { x ^ { 2 / 5 } + 2 }∫1∞​x2/5+2dx​

Correct AnswerverifiedShow Answer

Use reduction formulas to evaluate the integral. - ∫sin⁡23xcos⁡23xdx\int \sin ^ { 2 } 3 x \cos ^ { 2 } 3 x d x∫sin23xcos23xdx

Correct AnswerverifiedShow Answer

Solve the problem. -Find an upper bound for ∣ET∣\left| \mathrm { E } _ { \mathrm { T } } \right|∣ET​∣ in estimating ∫15(3x2+8) dx\int _ { 1 } ^ { 5 } \left( 3 \mathrm { x } ^ { 2 } + 8 \right) \mathrm { dx }∫15​(3x2+8) dx with n=6\mathrm { n } = 6n=6 steps.

Correct AnswerverifiedShow Answer

Expand the quotient by partial fractions. - x+4x2+2x+1\frac { x + 4 } { x ^ { 2 } + 2 x + 1 }x2+2x+1x+4​

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Solve the initial value problem for y as a function of x. - xdydx=x2−9,x≥3,y(3) =0x \frac { d y } { d x } = \sqrt { x ^ { 2 } - 9 } , x \geq 3 , y ( 3 ) = 0xdxdy​=x2−9​,x≥3,y(3) =0

Correct AnswerverifiedShow Answer

Use reduction formulas to evaluate the integral. - ∫8cos⁡32xdx\int 8 \cos ^ { 3 } 2 x d x∫8cos32xdx

Correct AnswerverifiedShow Answer

Evaluate the integral. - ∫1xln⁡x6dx\int \frac { 1 } { x \ln x ^ { 6 } } d x∫xlnx61​dx

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Provide an appropriate response. -(a) Find the values of ppp for which ∫011xpdx\int _ { 0 } ^ { 1 } \frac { 1 } { x p } d x∫01​xp1​dx converges. (b) Find the values of ppp for which ∫011xpdx\int _ { 0 } ^ { 1 } \frac { 1 } { x _ { p } } d x∫01​xp​1​dx diverges.

Correct Answerverified(a) The integral con...View AnswerShow Answer

Provide an appropriate response. - The "Simpson" sum is based upon the area under a ?‾. \text { The "Simpson" sum is based upon the area under a } \underline { ? } \text {. } The "Simpson" sum is based upon the area under a ?​.

Correct AnswerverifiedShow Answer

Use various trigonometric identities to simplify the expression then integrate. - ∫cos⁡7θsin⁡2θdθ\int \cos ^ { 7 } \theta \sin 2 \theta d \theta∫cos7θsin2θdθ

Correct AnswerverifiedShow Answer

Evaluate the improper integral. - $\int _ { 0 } ^ { 25 } \frac { d x } { \sqrt { | x - 16 | } }$

Correct Answer
verified

Find the value of the constant k so that the given function in a probability density function for a random variable over the specified interval. - $f ( x ) = k ^ { x }$ over $[ 0,20 ]$

Correct Answer
verified

Solve the initial value problem for y as a function of x. - $\sqrt { x ^ { 2 } - 4 } \frac { d y } { d x } = x , x > 2 , y ( 4 ) = 0$

Correct Answer
verified

Correct Answer
verified

Integrate the function. - $\int \frac { d x } { \sqrt { 25 x ^ { 2 } - 121 } } , x > \frac { 11 } { 5 }$

Correct Answer
verified

Determine whether the improper integral converges or diverges. - $\int _ { - \infty } ^ { \infty } \frac { d x } { \sqrt { 5 x ^ { 6 } + 1 } }$

Correct Answer
verified

Evaluate the integral by making a substitution (possibly trigonometric) and then applying a reduction formula. - $\int \frac { d x } { \left( 25 - x ^ { 2 } \right) ^ { 2 } }$

Correct Answer
verified

Use Simpson's Rule with n = 4 steps to estimate the integral. - $\int _ { - 1 } ^ { 0 } \sin \pi \mathrm { tdt }$

Correct Answer
verified

Determine whether the improper integral converges or diverges. - $\int _ { 1 } ^ { \infty } \frac { d x } { x ^ { 2 / 5 } + 2 }$

Correct Answer
verified

Use reduction formulas to evaluate the integral. - $\int \sin ^ { 2 } 3 x \cos ^ { 2 } 3 x d x$

Correct Answer
verified

Solve the problem. -Find an upper bound for $\left| \mathrm { E } _ { \mathrm { T } } \right|$ in estimating $\int _ { 1 } ^ { 5 } \left( 3 \mathrm { x } ^ { 2 } + 8 \right) \mathrm { dx }$ with $\mathrm { n } = 6$ steps.

Correct Answer
verified

Expand the quotient by partial fractions. - $\frac { x + 4 } { x ^ { 2 } + 2 x + 1 }$

Correct Answer
verified

Correct Answer
verified

Solve the initial value problem for y as a function of x. - $x \frac { d y } { d x } = \sqrt { x ^ { 2 } - 9 } , x \geq 3 , y ( 3 ) = 0$

Correct Answer
verified

Use reduction formulas to evaluate the integral. - $\int 8 \cos ^ { 3 } 2 x d x$

Correct Answer
verified

Evaluate the integral. - $\int \frac { 1 } { x \ln x ^ { 6 } } d x$

Correct Answer
verified

Correct Answer
verified

Provide an appropriate response. -(a) Find the values of $p$ for which $\int _ { 0 } ^ { 1 } \frac { 1 } { x p } d x$ converges. (b) Find the values of $p$ for which $\int _ { 0 } ^ { 1 } \frac { 1 } { x _ { p } } d x$ diverges.

Correct Answer
verified
(a) The integral con...
View Answer

Provide an appropriate response. - $\text { The "Simpson" sum is based upon the area under a } \underline { ? } \text {. }$

Correct Answer
verified

Use various trigonometric identities to simplify the expression then integrate. - $\int \cos ^ { 7 } \theta \sin 2 \theta d \theta$

Correct Answer
verified