Question 1

Solve the problem.
-The following table shows the rate of water flow (in gal/min)  from a stream into a pond during a 30 -minute period after a thunderstorm. Use Simpson's Rule to estimate the total amount of water flowing into the pond during this period. Round your answer to the nearest gallon.
  Time (min)   Rate (gal/min)  020052501030015250202202520030150\begin{array}{l|l}	ext { Time (min)  } & 	ext { Rate (gal/min)  } \\hline 0 & 200 \5 & 250 \10 & 300 \15 & 250 \20 & 220 \25 & 200 \30 & 150\end{array} Time (min)  051015202530​ Rate (gal/min)  200250300250220200150​​

Accepted Answer

A)   7975 gal
B)   6383 gal
C)   6983 gal
D)   6975 gal E) A) and B)
F) A) and C)

Question 2

Solve the problem.
-Find an upper bound for  ∣ES∣| \mathrm { E } _S |∣ES​∣  in estimating  ∫05(6x2−8) dx\int _ { 0 } ^ { 5 } \left( 6 \mathrm { x } ^ { 2 } - 8 \right)  \mathrm { dx }∫05​(6x2−8) dx  with  n=8\mathrm { n } = 8n=8  steps.

Accepted Answer

A)   0
B)    62512288\frac { 625 } { 12288 }12288625​ 
C)    62573728\frac { 625 } { 73728 }73728625​ 
D)    62524576\frac { 625 } { 24576 }24576625​E) A) and C)
F) A) and B)

Question 3

Integrate the function.
- ∫16−x2dx\int \sqrt { 16 - x ^ { 2 } } d x∫16−x2​dx

Accepted Answer

A)    8sin⁡−1(x4) +x16−x22+C8 \sin ^ { - 1 } \left( \frac { x } { 4 } \right)  + \frac { x \sqrt { 16 - x ^ { 2 } } } { 2 } + C8sin−1(4x​) +2x16−x2c-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 
B)    x1616−x2+16−x2x+C\frac { x } { 16 \sqrt { 16 - x ^ { 2 } } } + \frac { \sqrt { 16 - x ^ { 2 } } } { x } + C1616−x2c-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​+x16−x2c-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 
C)    16x16−x2+x2+C\frac { 16 x } { \sqrt { 16 - x ^ { 2 } } } + \frac { x } { 2 } + C16−x2c-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'/>​16x​+2x​+C 
D)    8x−x16−x22+C8 x - \frac { x \sqrt { 16 - x ^ { 2 } } } { 2 } + C8x−2x16−x2c-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'/>​​+CE) B) and C)
F) A) and B)

Question 4

Determine whether the improper integral converges or diverges.
- ∫1∞6x+2x2\int _ { 1 } ^ { \infty } \frac { \sqrt { 6 x + 2 } } { x ^ { 2 } }∫1∞​x26x+2​​

Accepted Answer

A)   Converges
B)   Diverges C) A) and B)
D) undefined

Question 5

Evaluate the integral.
- ∫x+10x+2dx\int \frac { x + 10 } { x + 2 } d x∫x+2x+10​dx

Accepted Answer

A)    x+8ln⁡∣x+2∣+Cx + 8 \ln | x + 2 | + Cx+8ln∣x+2∣+C 
B)    x+12ln⁡∣x+2∣+Cx + 12 \ln | x + 2 | + Cx+12ln∣x+2∣+C 
C)    −8(x+2) 2+C- \frac { 8 } { ( x + 2 )  ^ { 2 } } + C−(x+2) 28​+C 
D)    (x2/2+10x) ln⁡∣x+2∣+C\left( x ^ { 2 } / 2 + 10 x \right)  \ln | x + 2 | + C(x2/2+10x) ln∣x+2∣+CE) A) and B)
F) A) and C)

Question 6

Use a trigonometric substitution to evaluate the integral.
- ∫dxx(1+25ln⁡2x) \int \frac { d x } { x \left( 1 + 25 \ln ^ { 2 } x \right)  }∫x(1+25ln2x) dx​

Accepted Answer

A)    15xtan⁡−1(5ln⁡x) +C\frac { 1 } { 5 x } 	an ^ { - 1 } ( 5 \ln x )  + C5x1​tan−1(5lnx) +C 
B)    15tan⁡−1(25ln⁡2x) +C\frac { 1 } { 5 } 	an ^ { - 1 } \left( 25 \ln ^ { 2 } x ight)  + C51​tan−1(25ln2x) +C 
C)    150ln⁡(1+25ln⁡2x) +C\frac { 1 } { 50 } \ln \left( 1 + 25 \ln ^ { 2 } x ight)  + C501​ln(1+25ln2x) +C 
D)    15tan⁡−1(5ln⁡x) +C\frac { 1 } { 5 } 	an ^ { - 1 } ( 5 \ln x )  + C51​tan−1(5lnx) +CE) All of the above
F) B) and C)

Question 7

Determine whether the improper integral converges or diverges.
- ∫45e−x−4x−4dx\int _ { 4 } ^ { 5 } \frac { e ^ { - \sqrt { x - 4 } } } { \sqrt { x - 4 } } d x∫45​x−4​e−x−4​​dx

Accepted Answer

A)   Diverges
B)   Converges C) A) and B)
D) undefined

Question 8

Determine whether the function is a probability density function over the given interval.
-$$f ( x ) = 5 \sin 5 x \text { over } \left[ \frac { \pi } { 10 } , \frac { \pi } { 5 } \right]$$

Accepted Answer

A) True 
 B)False

Question 9

Solve the problem.
-Find the length of the curve  y=36−x2y = \sqrt { 36 - x ^ { 2 } }y=36−x2​  between  x=0x = 0x=0  and  x=3x = 3x=3 .

Accepted Answer

A)    16π\frac { 1 } { 6 } \pi61​π 
B)    2π2 \pi2π 
C)    32π\frac { 3 } { 2 } \pi23​π 
D)    π\piπE) A) and B)
F) A) and C)

Question 10

Find the area or volume.
-Find the volume of the solid generated by revolving the region under the curve  y=8e−2x\mathrm { y } = 8 \mathrm { e } ^ { - 2 \mathrm { x } }y=8e−2x  in the first quadrant about the  yyy -axis.

Accepted Answer

A)    4π4 \pi4π 
B)    64π64 \pi64π 
C)    8π8 \pi8π 
D)    8π38 \pi ^ { 3 }8π3E) All of the above
F) None of the above

Question 11

Evaluate the integral.
- ∫x2ln⁡9xdx\int x ^ { 2 } \ln 9 x d x∫x2ln9xdx

Accepted Answer

A)    13x3ln⁡9x−112x4+C\frac { 1 } { 3 } x ^ { 3 } \ln 9 x - \frac { 1 } { 12 } x ^ { 4 } + C31​x3ln9x−121​x4+C 
B)    ln⁡9x−13x3+C\ln 9 x - \frac { 1 } { 3 } x ^ { 3 } + Cln9x−31​x3+C 
C)    13x3ln⁡9x−19x3+C\frac { 1 } { 3 } x ^ { 3 } \ln 9 x - \frac { 1 } { 9 } x ^ { 3 } + C31​x3ln9x−91​x3+C 
D)    13x3ln⁡9x+19x3+C\frac { 1 } { 3 } x ^ { 3 } \ln 9 x + \frac { 1 } { 9 } x ^ { 3 } + C31​x3ln9x+91​x3+CE) B) and D)
F) All of the above

Question 12

Expand the quotient by partial fractions.
- 5x+3(x−3) (x−1) \frac { 5 x + 3 } { ( x - 3 )  ( x - 1 )  }(x−3) (x−1) 5x+3​

Accepted Answer

A)    9x−3+−4(x−3) (x−1) \frac { 9 } { x - 3 } + \frac { - 4 } { ( x - 3 )  ( x - 1 )  }x−39​+(x−3) (x−1) −4​ 
B)    9x−3+−4x−1\frac { 9 } { x - 3 } + \frac { - 4 } { x - 1 }x−39​+x−1−4​ 
C)    18x−3+8x−1\frac { 18 } { x - 3 } + \frac { 8 } { x - 1 }x−318​+x−18​ 
D)    9x−3+4x−1\frac { 9 } { x - 3 } + \frac { 4 } { x - 1 }x−39​+x−14​E) A) and C)
F) All of the above

Question 13

Use the Trapezoidal Rule with n = 4 steps to estimate the integral.
- ∫09xdx\int _ { 0 } ^ { 9 } x d x∫09​xdx

Accepted Answer

A)    814\frac { 81 } { 4 }481​ 
B)    812\frac { 81 } { 2 }281​ 
C)   81
D)    4058\frac { 405 } { 8 }8405​E) C) and D)
F) A) and B)

Question 14

Solve the problem.
-A rectangular swimming pool is being constructed, 18 feet long and 100 feet wide. The depth of the pool is measured at 3 -foot intervals across the length of the pool. Estimate the volume of water in the pool using the Trapezoidal Rule.
  Width (ft)   Depth (ft)  0535.56697127.5158189\begin{array}{l|l}	ext { Width (ft)  } & 	ext { Depth (ft)  } \\hline 0 & 5 \3 & 5.5 \6 & 6 \9 & 7 \12 & 7.5 \15 & 8 \18 & 9\end{array} Width (ft)  0369121518​ Depth (ft)  55.5677.589​​

Accepted Answer

A)    12,300ft312,300 \mathrm { ft } ^ { 3 }12,300ft3 
B)    8200ft38200 \mathrm { ft } ^ { 3 }8200ft3 
C)    14,400ft314,400 \mathrm { ft } ^ { 3 }14,400ft3 
D)    10,900ft310,900 \mathrm { ft } ^ { 3 }10,900ft3E) All of the above
F) C) and D)

Question 15

Find the area or volume.
-Find the volume of the solid generated by revolving the region under the curve  y=7xy = \frac { 7 } { x }y=x7​ , from  x=1x = 1x=1  to  x=∞x = \inftyx=∞ , about the  xxx -axis.

Accepted Answer

A)    7π7 \pi7π 
B)    17π\frac { 1 } { 7 } \pi71​π 
C)   7
D)    14π14 \pi14πE) A) and B)
F) A) and C)

Question 16

Use the Trapezoidal Rule with n = 4 steps to estimate the integral.
- ∫131x2dx\int _ { 1 } ^ { 3 } \frac { 1 } { x ^ { 2 } } d x∫13​x21​dx

Accepted Answer

A)    141200\frac { 141 } { 200 }200141​ 
B)    141400\frac { 141 } { 400 }400141​ 
C)    71100\frac { 71 } { 100 }10071​ 
D)    141100\frac { 141 } { 100 }100141​E) None of the above
F) A) and C)

Question 17

Find the indicated probability.
- f(x) =14e−x/4;[0,∞) ,P(1≤x≤4) \mathrm { f } ( \mathrm { x } )  = \frac { 1 } { 4 } \mathrm { e } ^ { - \mathrm { x } / 4 } ; [ 0 , \infty )  , \mathrm { P } ( 1 \leq \mathrm { x } \leq 4 ) f(x) =41​e−x/4;[0,∞) ,P(1≤x≤4)

Accepted Answer

A)   0.0514
B)   0.1027
C)   0.0685
D)   0.4109 E) All of the above
F) C) and D)

Question 18

Integrate the function.
- ∫9dx9+x2\int \frac { 9 \mathrm { dx } } { \sqrt { 9 + \mathrm { x } ^ { 2 } } }∫9+x2​9dx​

Accepted Answer

A)    9ln⁡∣x+9+x2∣+C9 \ln \left| x + \sqrt { 9 + x ^ { 2 } } \right| + C9lnc10,0,16.667,-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+9+x2c-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 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)    2x2+9+C\frac { 2 } { x ^ { 2 } + 9 } + Cx2+92​+C 
C)    9ln⁡∣9+x2∣+C9 \ln \left| \sqrt { 9 + x ^ { 2 } } \right| + C9lnc10,0,16.667,-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'/>​9+x2c-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 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)    x+ln⁡∣9+9+x2∣+Cx + \ln \left| 9 + \sqrt { 9 + x ^ { 2 } } \right| + Cx+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'/>​9+9+x2c-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 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'/>​+CE) B) and D)
F) A) and C)

Question 19

Evaluate the integral.
- ∫y3e−7ydy\int y ^ { 3 } e ^ { - 7 y } d y∫y3e−7ydy

Accepted Answer

A)    −17e−7y[y3+y2+y+6]+C- \frac { 1 } { 7 } \mathrm { e } ^ { - 7 \mathrm { y } } \left[ \mathrm { y } ^ { 3 } + \mathrm { y } ^ { 2 } + \mathrm { y } + 6 \right] + C−71​e−7y[y3+y2+y+6]+C 
B)    −128y4e−7y+C- \frac { 1 } { 28 } y ^ { 4 } e ^ { - 7 y } + C−281​y4e−7y+C 
C)    −e−7y[17y3+349y2+6343y+62401]+C- \mathrm { e } ^ { - 7 \mathrm { y } } \left[ \frac { 1 } { 7 } \mathrm { y } ^ { 3 } + \frac { 3 } { 49 } \mathrm { y } ^ { 2 } + \frac { 6 } { 343 } \mathrm { y } + \frac { 6 } { 2401 } \right] + \mathrm { C }−e−7y[71​y3+493​y2+3436​y+24016​]+C 
D)    e−7y[17y3−349y2+6343y−62401]+Ce ^ { - 7 y } \left[ \frac { 1 } { 7 } y ^ { 3 } - \frac { 3 } { 49 } y ^ { 2 } + \frac { 6 } { 343 } y - \frac { 6 } { 2401 } \right] + Ce−7y[71​y3−493​y2+3436​y−24016​]+CE) B) and C)
F) A) and D)

Question 20

Evaluate the integral.
- ∫1x49+x2dx\int \frac { 1 } { x \sqrt { 49 + x ^ { 2 } } } d x∫x49+x2​1​dx

Accepted Answer

A)    −17ln⁡∣7+49+x2x2∣+C- \frac { 1 } { 7 } \ln \left| \frac { 7 + \sqrt { 49 + x ^ { 2 } } } { x ^ { 2 } } \right| + C−71​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'/>​x27+49+x2c-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)    −17ln⁡∣7−49+x2x∣+C- \frac { 1 } { 7 } \ln \left| \frac { 7 - \sqrt { 49 + x ^ { 2 } } } { x } \right| + C−71​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'/>​x7−49+x2c-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)    −17ln⁡∣7+49+x2x∣+C- \frac { 1 } { 7 } \ln \left| \frac { 7 + \sqrt { 49 + x ^ { 2 } } } { x } \right| + C−71​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'/>​x7+49+x2c-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)    17ln⁡∣7+49+x2x∣+C\frac { 1 } { 7 } \ln \left| \frac { 7 + \sqrt { 49 + \mathrm { x } ^ { 2 } } } { \mathrm { x } } \right| + C71​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'/>​x7+49+x2c-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'/>​+CE) B) and D)
F) B) and C)

Exam 9: Techniques of Integration

Determine whether the improper integral converges or diverges. - $\int _ { 1 } ^ { \infty } \frac { \sqrt { 6 x + 2 } } { x ^ { 2 } }$

Correct Answer
verified

Find the area or volume. -Find the volume of the solid generated by revolving the region under the curve $\mathrm { y } = 8 \mathrm { e } ^ { - 2 \mathrm { x } }$ in the first quadrant about the $y$ -axis.

Correct Answer
verified

Evaluate the integral. - $\int \frac { x + 10 } { x + 2 } d x$

Correct Answer
verified

Integrate the function. - $\int \frac { 9 \mathrm { dx } } { \sqrt { 9 + \mathrm { x } ^ { 2 } } }$

Correct Answer
verified

Use the Trapezoidal Rule with n = 4 steps to estimate the integral. - $\int _ { 1 } ^ { 3 } \frac { 1 } { x ^ { 2 } } d x$

Correct Answer
verified

Correct Answer
verified

Integrate the function. - $\int \sqrt { 16 - x ^ { 2 } } d x$

Correct Answer
verified

Find the area or volume. -Find the volume of the solid generated by revolving the region under the curve $y = \frac { 7 } { x }$ , from $x = 1$ to $x = \infty$ , about the $x$ -axis.

Correct Answer
verified

Solve the problem. -Find the length of the curve $y = \sqrt { 36 - x ^ { 2 } }$ between $x = 0$ and $x = 3$ .

Correct Answer
verified

Evaluate the integral. - $\int \frac { 1 } { x \sqrt { 49 + x ^ { 2 } } } d x$

Correct Answer
verified

Determine whether the function is a probability density function over the given interval. - $f ( x ) = 5 \sin 5 x \text { over } \left[ \frac { \pi } { 10 } , \frac { \pi } { 5 } \right]$

Correct Answer
verified

Use a trigonometric substitution to evaluate the integral. - $\int \frac { d x } { x \left( 1 + 25 \ln ^ { 2 } x \right) }$

Correct Answer
verified

Solve the problem. -Find an upper bound for $| \mathrm { E } _S |$ in estimating $\int _ { 0 } ^ { 5 } \left( 6 \mathrm { x } ^ { 2 } - 8 \right) \mathrm { dx }$ with $\mathrm { n } = 8$ steps.

Correct Answer
verified

Find the indicated probability. - $\mathrm { f } ( \mathrm { x } ) = \frac { 1 } { 4 } \mathrm { e } ^ { - \mathrm { x } / 4 } ; [ 0 , \infty ) , \mathrm { P } ( 1 \leq \mathrm { x } \leq 4 )$

Correct Answer
verified

Determine whether the improper integral converges or diverges. - $\int _ { 4 } ^ { 5 } \frac { e ^ { - \sqrt { x - 4 } } } { \sqrt { x - 4 } } d x$

Correct Answer
verified

Expand the quotient by partial fractions. - $\frac { 5 x + 3 } { ( x - 3 ) ( x - 1 ) }$

Correct Answer
verified

Correct Answer
verified

Evaluate the integral. - $\int x ^ { 2 } \ln 9 x d x$

Correct Answer
verified

Evaluate the integral. - $\int y ^ { 3 } e ^ { - 7 y } d y$

Correct Answer
verified

Use the Trapezoidal Rule with n = 4 steps to estimate the integral. - $\int _ { 0 } ^ { 9 } x d x$

Correct Answer
verified

Exam 9: Techniques of Integration

Determine whether the improper integral converges or diverges. - ∫1∞6x+2x2\int _ { 1 } ^ { \infty } \frac { \sqrt { 6 x + 2 } } { x ^ { 2 } }∫1∞​x26x+2​​

Correct AnswerverifiedShow Answer

Find the area or volume. -Find the volume of the solid generated by revolving the region under the curve y=8e−2x\mathrm { y } = 8 \mathrm { e } ^ { - 2 \mathrm { x } }y=8e−2x in the first quadrant about the yyy -axis.

Correct AnswerverifiedShow Answer

Evaluate the integral. - ∫x+10x+2dx\int \frac { x + 10 } { x + 2 } d x∫x+2x+10​dx

Correct AnswerverifiedShow Answer

Integrate the function. - ∫9dx9+x2\int \frac { 9 \mathrm { dx } } { \sqrt { 9 + \mathrm { x } ^ { 2 } } }∫9+x2​9dx​

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Use the Trapezoidal Rule with n = 4 steps to estimate the integral. - ∫131x2dx\int _ { 1 } ^ { 3 } \frac { 1 } { x ^ { 2 } } d x∫13​x21​dx

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Integrate the function. - ∫16−x2dx\int \sqrt { 16 - x ^ { 2 } } d x∫16−x2​dx

Correct AnswerverifiedShow Answer

Find the area or volume. -Find the volume of the solid generated by revolving the region under the curve y=7xy = \frac { 7 } { x }y=x7​ , from x=1x = 1x=1 to x=∞x = \inftyx=∞ , about the xxx -axis.

Correct AnswerverifiedShow Answer

Solve the problem. -Find the length of the curve y=36−x2y = \sqrt { 36 - x ^ { 2 } }y=36−x2​ between x=0x = 0x=0 and x=3x = 3x=3 .

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Evaluate the integral. - ∫1x49+x2dx\int \frac { 1 } { x \sqrt { 49 + x ^ { 2 } } } d x∫x49+x2​1​dx

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Determine whether the function is a probability density function over the given interval. - f(x)=5sin⁡5x over [π10,π5]f ( x ) = 5 \sin 5 x \text { over } \left[ \frac { \pi } { 10 } , \frac { \pi } { 5 } \right]f(x)=5sin5x over [10π​,5π​]

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Use a trigonometric substitution to evaluate the integral. - ∫dxx(1+25ln⁡2x) \int \frac { d x } { x \left( 1 + 25 \ln ^ { 2 } x \right) }∫x(1+25ln2x) dx​

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Solve the problem. -Find an upper bound for ∣ES∣| \mathrm { E } _S |∣ES​∣ in estimating ∫05(6x2−8) dx\int _ { 0 } ^ { 5 } \left( 6 \mathrm { x } ^ { 2 } - 8 \right) \mathrm { dx }∫05​(6x2−8) dx with n=8\mathrm { n } = 8n=8 steps.

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Find the indicated probability. - f(x) =14e−x/4;[0,∞) ,P(1≤x≤4) \mathrm { f } ( \mathrm { x } ) = \frac { 1 } { 4 } \mathrm { e } ^ { - \mathrm { x } / 4 } ; [ 0 , \infty ) , \mathrm { P } ( 1 \leq \mathrm { x } \leq 4 ) f(x) =41​e−x/4;[0,∞) ,P(1≤x≤4)

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Determine whether the improper integral converges or diverges. - ∫45e−x−4x−4dx\int _ { 4 } ^ { 5 } \frac { e ^ { - \sqrt { x - 4 } } } { \sqrt { x - 4 } } d x∫45​x−4​e−x−4​​dx

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Expand the quotient by partial fractions. - 5x+3(x−3) (x−1) \frac { 5 x + 3 } { ( x - 3 ) ( x - 1 ) }(x−3) (x−1) 5x+3​

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Evaluate the integral. - ∫x2ln⁡9xdx\int x ^ { 2 } \ln 9 x d x∫x2ln9xdx

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Evaluate the integral. - ∫y3e−7ydy\int y ^ { 3 } e ^ { - 7 y } d y∫y3e−7ydy

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Use the Trapezoidal Rule with n = 4 steps to estimate the integral. - ∫09xdx\int _ { 0 } ^ { 9 } x d x∫09​xdx

Correct AnswerverifiedShow Answer

Determine whether the improper integral converges or diverges. - $\int _ { 1 } ^ { \infty } \frac { \sqrt { 6 x + 2 } } { x ^ { 2 } }$

Correct Answer
verified

Find the area or volume. -Find the volume of the solid generated by revolving the region under the curve $\mathrm { y } = 8 \mathrm { e } ^ { - 2 \mathrm { x } }$ in the first quadrant about the $y$ -axis.

Correct Answer
verified

Evaluate the integral. - $\int \frac { x + 10 } { x + 2 } d x$

Correct Answer
verified

Integrate the function. - $\int \frac { 9 \mathrm { dx } } { \sqrt { 9 + \mathrm { x } ^ { 2 } } }$

Correct Answer
verified

Use the Trapezoidal Rule with n = 4 steps to estimate the integral. - $\int _ { 1 } ^ { 3 } \frac { 1 } { x ^ { 2 } } d x$

Correct Answer
verified

Correct Answer
verified

Integrate the function. - $\int \sqrt { 16 - x ^ { 2 } } d x$

Correct Answer
verified

Find the area or volume. -Find the volume of the solid generated by revolving the region under the curve $y = \frac { 7 } { x }$ , from $x = 1$ to $x = \infty$ , about the $x$ -axis.

Correct Answer
verified

Solve the problem. -Find the length of the curve $y = \sqrt { 36 - x ^ { 2 } }$ between $x = 0$ and $x = 3$ .

Correct Answer
verified

Evaluate the integral. - $\int \frac { 1 } { x \sqrt { 49 + x ^ { 2 } } } d x$

Correct Answer
verified

Determine whether the function is a probability density function over the given interval. - $f ( x ) = 5 \sin 5 x \text { over } \left[ \frac { \pi } { 10 } , \frac { \pi } { 5 } \right]$

Correct Answer
verified

Use a trigonometric substitution to evaluate the integral. - $\int \frac { d x } { x \left( 1 + 25 \ln ^ { 2 } x \right) }$

Correct Answer
verified

Solve the problem. -Find an upper bound for $| \mathrm { E } _S |$ in estimating $\int _ { 0 } ^ { 5 } \left( 6 \mathrm { x } ^ { 2 } - 8 \right) \mathrm { dx }$ with $\mathrm { n } = 8$ steps.

Correct Answer
verified

Find the indicated probability. - $\mathrm { f } ( \mathrm { x } ) = \frac { 1 } { 4 } \mathrm { e } ^ { - \mathrm { x } / 4 } ; [ 0 , \infty ) , \mathrm { P } ( 1 \leq \mathrm { x } \leq 4 )$

Correct Answer
verified

Determine whether the improper integral converges or diverges. - $\int _ { 4 } ^ { 5 } \frac { e ^ { - \sqrt { x - 4 } } } { \sqrt { x - 4 } } d x$

Correct Answer
verified

Expand the quotient by partial fractions. - $\frac { 5 x + 3 } { ( x - 3 ) ( x - 1 ) }$

Correct Answer
verified

Correct Answer
verified

Evaluate the integral. - $\int x ^ { 2 } \ln 9 x d x$

Correct Answer
verified

Evaluate the integral. - $\int y ^ { 3 } e ^ { - 7 y } d y$

Correct Answer
verified

Use the Trapezoidal Rule with n = 4 steps to estimate the integral. - $\int _ { 0 } ^ { 9 } x d x$

Correct Answer
verified