Question 1

The average value of  f(x) =1+sin⁡xf ( x )  = 1 + \sin xf(x) =1+sinx  on  [0,π][ 0 , \pi ][0,π]  is

Accepted Answer

A)    2+π2\frac { 2 + \pi } { 2 }22+π​ 
B)    2−π2\frac { 2 - \pi } { 2 }22−π​ 
C)    π−22\frac { \pi - 2 } { 2 }2π−2​ 
D)    2+ππ\frac { 2 + \pi } { \pi }π2+π​ 
E)    2−ππ\frac { 2 - \pi } { \pi }π2−π​F) A) and D)
G) None of the above

Question 2

The indefinite integral  ∫x2ex3−2dx\int x ^ { 2 } e ^ { x ^ { 3 } - 2 } d x∫x2ex3−2dx  is

Accepted Answer

A)    x3ex33+C\frac { x ^ { 3 } e ^ { x ^ { 3 } } } { 3 } + C3x3ex3​+C 
B)    6xex3+C6 x e ^ { x ^ { 3 } } + C6xex3+C 
C)    ex3−23+C\frac { e ^ { x ^ { 3 } - 2 } } { 3 } + C3ex3−2​+C 
D)    ex3−22+C\frac { e ^ { x ^ { 3 } - 2 } } { 2 } + C2ex3−2​+C 
E)    2x3ex3−23+C\frac { 2 x ^ { 3 } e ^ { x ^ { 3 } - 2 } } { 3 } + C32x3ex3−2​+CF) None of the above
G) D) and E)

Question 3

Let A denote the area enclosed by the graph  f(x) =1xf ( x )  = \frac { 1 } { x }f(x) =x1​  , the x-axis, and the lines x = 1 and x = e. By part 2 of the Fundamental Theorem of Calculus, A is

Accepted Answer

A)    1−1e1 - \frac { 1 } { \mathrm { e } }1−e1​ 
B)    1−1e21 - \frac { 1 } { \mathrm { e } ^ { 2 } }1−e21​ 
C)    12\frac { 1 } { 2 }21​ 
D)  1
E)    1e2\frac { 1 } { \mathrm { e } ^ { 2 } }e21​F) C) and D)
G) A) and B)

Question 4

The derivative  ddx[∫0sin⁡x1−t2dt]\frac { d } { d x } \left[ \int _ { 0 } ^ { \sin x } \sqrt { 1 - t ^ { 2 } } d t \right]dxd​[∫0sinx​1−t2​dt]  is

Accepted Answer

A)  cos x
B)  cos2 x
C)  sec x
D)  sec2 x
E)  csc2 x F) All of the above
G) A) and C) B

Question 5

If f is an even function,  ∫−40f(x) dx=4,\int _ { - 4 } ^ { 0 } f ( x )  d x = 4,∫−40​f(x) dx=4,  and  ∫02f(x) dx=5,\int _ { 0 } ^ { 2 } f ( x )  d x = 5,∫02​f(x) dx=5,  then  ∫24f(x) dx\int _ { 2 } ^ { 4 } f ( x )  d x∫24​f(x) dx  is

Accepted Answer

A)    −3- 3−3 
B)    −2- 2−2 
C)    −1- 1−1 
D)  0
E)    111F) A) and E)
G) A) and C)

Question 6

Let A denote the area enclosed by the graph  f(x) =∣x∣, f ( x )  = | x | 	ext {, }f(x) =∣x∣,   the x-axis, and the lines  x=−1x = - 1x=−1  and  x=1x = 1x=1  . Graphing the region and using plane geometry, we can find that A is

Accepted Answer

A)  0.5
B)  1
C)  1.5
D)  2
E)  2.5 F) D) and E)
G) B) and C)

Question 7

The solution to the differential equation  dydx=ex2y\frac { d y } { d x } = \frac { e ^ { x } } { 2 y }dxdy​=2yex​  satisfying the boundary condition  y=1y = 1y=1  when  x=0x = 0x=0  is

Accepted Answer

A)    y2=exy ^ { 2 } = e ^ { x }y2=ex 
B)    y2=ex+1y ^ { 2 } = e ^ { x } + 1y2=ex+1 
C)    y2=ex−1y ^ { 2 } = e ^ { x } - 1y2=ex−1 
D)    y2=ex+3y ^ { 2 } = e ^ { x } + 3y2=ex+3 
E)    y2=ex+2y ^ { 2 } = e ^ { x } + 2y2=ex+2F) C) and E)
G) A) and B) A

Question 8

The solution to the differential equation  dydx=yx\frac { d y } { d x } = \frac { y } { x }dxdy​=xy​  satisfying the boundary condition  y=1y = 1y=1  when  x=1x = 1x=1  is

Accepted Answer

A)    y=xy = xy=x 
B)    y=x−1y = x - 1y=x−1 
C)    y=exy = e ^ { x }y=ex 
D)    y=x+1y = x + 1y=x+1 
E)    y=ex+1y = e ^ { x + 1 }y=ex+1F) B) and C)
G) A) and E)

Question 9

The rate of water consumption (in hundreds of gallons per year)  in an office building since its opening in 1995 is modeled by the function  w=14tw = \frac { 1 } { 4 } tw=41​t  , where tis the number of years after 1995. Which integral represents the total number of gallons consumed between 1996 and 2001?

Accepted Answer

A)    ∫0614tdt\int _ { 0 } ^ { 6 } \frac { 1 } { 4 } t d t∫06​41​tdt 
B)    ∫0514tdt\int _ { 0 } ^ { 5 } \frac { 1 } { 4 } t d t∫05​41​tdt 
C)    ∫1614tdt\int _ { 1 } ^ { 6 } \frac { 1 } { 4 } t d t∫16​41​tdt 
D)    ∫1614dt\int _ { 1 } ^ { 6 } \frac { 1 } { 4 } d t∫16​41​dt 
E)    ∫0514dt\int _ { 0 } ^ { 5 } \frac { 1 } { 4 } d t∫05​41​dtF) A) and C)
G) B) and D)

Question 10

The derivative  ddx[∫x32tdt]\frac { d } { d x } \left[ \int _ { x } ^ { 3 } 2 ^ { t } d t \right]dxd​[∫x3​2tdt]  is

Accepted Answer

A)    2x+2−82 ^ { x + 2 } - 82x+2−8 
B)    8−2x+18 - 2 ^ { x + 1 }8−2x+1 
C)    −2x+1- 2 ^ { x + 1 }−2x+1 
D)    8−2x+1x+18 - \frac { 2 ^ { x + 1 } } { x + 1 }8−x+12x+1​ 
E)    −2x- 2 ^ { x }−2xF) B) and D)
G) C) and E)

Question 11

The solution to the differential equation  dydx=2xy\frac { d y } { d x } = 2 x ydxdy​=2xy  satisfying the boundary condition  y=e2y = e ^ { 2 }y=e2  when  x=0x = 0x=0  is

Accepted Answer

A)    y=ex2y = e ^ { x ^ { 2 } }y=ex2 
B)    y=ex2−1y = e ^ { x ^ { 2 } - 1 }y=ex2−1 
C)    y=ex2+1y = e ^ { x ^ { 2 } + 1 }y=ex2+1 
D)    y=ex2+2y = e ^ { x ^ { 2 } + 2 }y=ex2+2 
E)    y=ex2+3y = e ^ { x ^ { 2 } + 3 }y=ex2+3F) None of the above
G) A) and E)

Question 12

The derivative  ddx[∫3x2+12tdt]\frac { d } { d x } \left[ \int _ { 3 } ^ { x ^ { 2 } + 1 } 2 ^ { t } d t \right]dxd​[∫3x2+1​2tdt]  is

Accepted Answer

A)    2x2+12 ^ { x ^ { 2 } + 1 }2x2+1 
B)    2x2+1(x2+1) 2 ^ { x ^ { 2 } + 1 } \left( x ^ { 2 } + 1 \right) 2x2+1(x2+1)  
C)    x(2x2+2) x \left( 2 ^ { x ^ { 2 } + 2 } \right) x(2x2+2)  
D)    x(2x2+1) −3x \left( 2 ^ { x ^ { 2 } + 1 } \right)  - 3x(2x2+1) −3 
E)    2x2+1−32 ^ { x ^ { 2 } + 1 } - 32x2+1−3F) B) and D)
G) None of the above

Question 13

The antiderivative  ∫(ex+xe) dx\int \left( e ^ { x } + x ^ { e } \right)  d x∫(ex+xe) dx  is

Accepted Answer

A)    ex+1x+1+xe+1e+1+C\frac { e ^ { x + 1 } } { x + 1 } + \frac { x ^ { e + 1 } } { e + 1 } + Cx+1ex+1​+e+1xe+1​+C 
B)    ex+ex+1e+1+C\frac { e ^ { x } + e ^ { x + 1 } } { e + 1 } + Ce+1ex+ex+1​+C 
C)    ex+xe+1e+1+Ce ^ { x } + \frac { x ^ { e + 1 } } { e + 1 } + Cex+e+1xe+1​+C 
D)    ex+exe−1+Ce ^ { x } + e x ^ { e - 1 } + Cex+exe−1+C 
E)    ex+xe+Ce ^ { x } + x ^ { e } + Cex+xe+CF) D) and E)
G) None of the above

Question 14

If  F(x) =∫1x(t+1) dtF ( x )  = \int _ { 1 } ^ { x } ( \sqrt { t } + 1 )  d tF(x) =∫1x​(t​+1) dt  , what is F(4) ?

Accepted Answer

A)  1
B)    313\frac { 31 } { 3 }331​ 
C)  12
D)    233\frac { 23 } { 3 }323​ 
E)    283\frac { 28 } { 3 }328​F) A) and E)
G) B) and D)

Question 15

Let  ∫13f(x) dx=14\int _ { 1 } ^ { 3 } f ( x )  d x = 14∫13​f(x) dx=14  and  ∫83f(x) dx=−5\int _ { 8 } ^ { 3 } f ( x )  d x = - 5∫83​f(x) dx=−5  Then  ∫18f(x) dx\int _ { 1 } ^ { 8 } f ( x )  d x∫18​f(x) dx  is

Accepted Answer

A)    −19- 19−19 
B)    −9- 9−9 
C)  9
D)  14
E)  19 F) B) and D)
G) A) and E)

Question 16

The bounds m and M used in the Bounds on an Integral Theorem for  ∫−12[2−(x−1) 2]dx\int _ { - 1 } ^ { 2 } \left[ 2 - ( x - 1 )  ^ { 2 } \right] d x∫−12​[2−(x−1) 2]dx  are

Accepted Answer

A)    −1,2- 1,2−1,2 
B)    −2,1- 2,1−2,1 
C)    −2,−1- 2 , - 1−2,−1 
D)    −2,2- 2,2−2,2 
E)    −2,3- 2,3−2,3F) A) and E)
G) A) and D)

Question 17

The derivative  ddx[∫x2+14ln⁡(1t) dt]\frac { d } { d x } \left[ \int _ { x ^ { 2 } + 1 } ^ { 4 } \ln \left( \frac { 1 } { t } \right)  d t \right]dxd​[∫x2+14​ln(t1​) dt]  is

Accepted Answer

A)    2xln⁡(x2+1) 2 x \ln \left( x ^ { 2 } + 1 \right) 2xln(x2+1)  
B)    2xln⁡(1x2+1) 2 x \ln \left( \frac { 1 } { x ^ { 2 } + 1 } \right) 2xln(x2+11​)  
C)    2xln⁡(x2+1) −ln⁡(14) 2 x \ln \left( x ^ { 2 } + 1 \right)  - \ln \left( \frac { 1 } { 4 } \right) 2xln(x2+1) −ln(41​)  
D)    2xln⁡(1x2+1) −ln⁡(14) 2 x \ln \left( \frac { 1 } { x ^ { 2 } + 1 } \right)  - \ln \left( \frac { 1 } { 4 } \right) 2xln(x2+11​) −ln(41​)  
E)    2xln⁡(x2+1) −ln⁡42 x \ln \left( x ^ { 2 } + 1 \right)  - \ln 42xln(x2+1) −ln4F) All of the above
G) A) and C)

Question 18

The antiderivative  ∫1xx3dx\int \frac { 1 } { x \sqrt [ 3 ] { x } } d x∫x3x​1​dx  is

Accepted Answer

A)    −1x3+C- \frac { 1 } { \sqrt [ 3 ] { x } } + C−3xc-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'/>​1​+C 
B)    −13x3+C- \frac { 1 } { 3 \sqrt [ 3 ] { x } } + C−33xc-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'/>​1​+C 
C)    −3x3+C- \frac { 3 } { \sqrt [ 3 ] { x } } + C−3xc-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'/>​3​+C 
D)    13x3+C\frac { 1 } { 3 \sqrt [ 3 ] { x } } + C33xc-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'/>​1​+C 
E)    3x3+C\frac { 3 } { \sqrt [ 3 ] { x } } + C3xc-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'/>​3​+CF) All of the above
G) A) and D)

Question 19

The average value of  f(x) =∣2x−3∣f ( x )  = | 2 x - 3 |f(x) =∣2x−3∣  on [0, 3] is

Accepted Answer

A)    32\frac { 3 } { 2 }23​ 
B)    53\frac { 5 } { 3 }35​ 
C)    73\frac { 7 } { 3 }37​ 
D)    93\frac { 9 } { 3 }39​ 
E)    92\frac { 9 } { 2 }29​F) B) and D)
G) A) and E) A

Question 20

Suppose S6 is the upper sum of the area enclosed by the graph  f(x) =10−x2,f ( x )  = 10 - x ^ { 2 },f(x) =10−x2,  the x-axis, and the lines x = 0 and x = 3 by partitioning [0, 3] into six subintervals [0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2], [2, 2.5], and [2.5, 3]. Then S6 is

Accepted Answer

A)  23.125
B)  23.225
C)  23.235
D)  23.245
E)  23.255 F) A) and C)
G) D) and E)

Exam 6: The Integral

The average value of f(x) =1+sin⁡xf ( x ) = 1 + \sin xf(x) =1+sinx on [0,π][ 0 , \pi ][0,π] is

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The indefinite integral ∫x2ex3−2dx\int x ^ { 2 } e ^ { x ^ { 3 } - 2 } d x∫x2ex3−2dx is

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Let A denote the area enclosed by the graph f(x) =1xf ( x ) = \frac { 1 } { x }f(x) =x1​ , the x-axis, and the lines x = 1 and x = e. By part 2 of the Fundamental Theorem of Calculus, A is

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The derivative ddx[∫0sin⁡x1−t2dt]\frac { d } { d x } \left[ \int _ { 0 } ^ { \sin x } \sqrt { 1 - t ^ { 2 } } d t \right]dxd​[∫0sinx​1−t2​dt] is

Correct AnswerverifiedB

If f is an even function, ∫−40f(x) dx=4,\int _ { - 4 } ^ { 0 } f ( x ) d x = 4,∫−40​f(x) dx=4, and ∫02f(x) dx=5,\int _ { 0 } ^ { 2 } f ( x ) d x = 5,∫02​f(x) dx=5, then ∫24f(x) dx\int _ { 2 } ^ { 4 } f ( x ) d x∫24​f(x) dx is

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Let A denote the area enclosed by the graph f(x) =∣x∣, f ( x ) = | x | \text {, }f(x) =∣x∣, the x-axis, and the lines x=−1x = - 1x=−1 and x=1x = 1x=1 . Graphing the region and using plane geometry, we can find that A is

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The solution to the differential equation dydx=ex2y\frac { d y } { d x } = \frac { e ^ { x } } { 2 y }dxdy​=2yex​ satisfying the boundary condition y=1y = 1y=1 when x=0x = 0x=0 is

Correct AnswerverifiedA

The solution to the differential equation dydx=yx\frac { d y } { d x } = \frac { y } { x }dxdy​=xy​ satisfying the boundary condition y=1y = 1y=1 when x=1x = 1x=1 is

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The derivative ddx[∫x32tdt]\frac { d } { d x } \left[ \int _ { x } ^ { 3 } 2 ^ { t } d t \right]dxd​[∫x3​2tdt] is

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The solution to the differential equation dydx=2xy\frac { d y } { d x } = 2 x ydxdy​=2xy satisfying the boundary condition y=e2y = e ^ { 2 }y=e2 when x=0x = 0x=0 is

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The derivative ddx[∫3x2+12tdt]\frac { d } { d x } \left[ \int _ { 3 } ^ { x ^ { 2 } + 1 } 2 ^ { t } d t \right]dxd​[∫3x2+1​2tdt] is

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The antiderivative ∫(ex+xe) dx\int \left( e ^ { x } + x ^ { e } \right) d x∫(ex+xe) dx is

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If F(x) =∫1x(t+1) dtF ( x ) = \int _ { 1 } ^ { x } ( \sqrt { t } + 1 ) d tF(x) =∫1x​(t​+1) dt , what is F(4) ?

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Let ∫13f(x) dx=14\int _ { 1 } ^ { 3 } f ( x ) d x = 14∫13​f(x) dx=14 and ∫83f(x) dx=−5\int _ { 8 } ^ { 3 } f ( x ) d x = - 5∫83​f(x) dx=−5 Then ∫18f(x) dx\int _ { 1 } ^ { 8 } f ( x ) d x∫18​f(x) dx is

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The bounds m and M used in the Bounds on an Integral Theorem for ∫−12[2−(x−1) 2]dx\int _ { - 1 } ^ { 2 } \left[ 2 - ( x - 1 ) ^ { 2 } \right] d x∫−12​[2−(x−1) 2]dx are

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The derivative ddx[∫x2+14ln⁡(1t) dt]\frac { d } { d x } \left[ \int _ { x ^ { 2 } + 1 } ^ { 4 } \ln \left( \frac { 1 } { t } \right) d t \right]dxd​[∫x2+14​ln(t1​) dt] is

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The antiderivative ∫1xx3dx\int \frac { 1 } { x \sqrt [ 3 ] { x } } d x∫x3x​1​dx is

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The average value of f(x) =∣2x−3∣f ( x ) = | 2 x - 3 |f(x) =∣2x−3∣ on [0, 3] is

Correct AnswerverifiedA

Suppose S6 is the upper sum of the area enclosed by the graph f(x) =10−x2,f ( x ) = 10 - x ^ { 2 },f(x) =10−x2, the x-axis, and the lines x = 0 and x = 3 by partitioning [0, 3] into six subintervals [0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2], [2, 2.5], and [2.5, 3]. Then S6 is

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The average value of $f ( x ) = 1 + \sin x$ on $[ 0 , \pi ]$ is

Correct Answer
verified

The indefinite integral $\int x ^ { 2 } e ^ { x ^ { 3 } - 2 } d x$ is

Correct Answer
verified

Let A denote the area enclosed by the graph $f ( x ) = \frac { 1 } { x }$ , the x-axis, and the lines x = 1 and x = e. By part 2 of the Fundamental Theorem of Calculus, A is

Correct Answer
verified

The derivative $\frac { d } { d x } \left[ \int _ { 0 } ^ { \sin x } \sqrt { 1 - t ^ { 2 } } d t \right]$ is

Correct Answer
verified
B

If f is an even function, $\int _ { - 4 } ^ { 0 } f ( x ) d x = 4,$ and $\int _ { 0 } ^ { 2 } f ( x ) d x = 5,$ then $\int _ { 2 } ^ { 4 } f ( x ) d x$ is

Correct Answer
verified

Let A denote the area enclosed by the graph $f ( x ) = | x | \text {, }$ the x-axis, and the lines $x = - 1$ and $x = 1$ . Graphing the region and using plane geometry, we can find that A is

Correct Answer
verified

The solution to the differential equation $\frac { d y } { d x } = \frac { e ^ { x } } { 2 y }$ satisfying the boundary condition $y = 1$ when $x = 0$ is

Correct Answer
verified
A

The solution to the differential equation $\frac { d y } { d x } = \frac { y } { x }$ satisfying the boundary condition $y = 1$ when $x = 1$ is

Correct Answer
verified

Correct Answer
verified

The derivative $\frac { d } { d x } \left[ \int _ { x } ^ { 3 } 2 ^ { t } d t \right]$ is

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The solution to the differential equation $\frac { d y } { d x } = 2 x y$ satisfying the boundary condition $y = e ^ { 2 }$ when $x = 0$ is

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The derivative $\frac { d } { d x } \left[ \int _ { 3 } ^ { x ^ { 2 } + 1 } 2 ^ { t } d t \right]$ is

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The antiderivative $\int \left( e ^ { x } + x ^ { e } \right) d x$ is

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If $F ( x ) = \int _ { 1 } ^ { x } ( \sqrt { t } + 1 ) d t$ , what is F(4) ?

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Let $\int _ { 1 } ^ { 3 } f ( x ) d x = 14$ and $\int _ { 8 } ^ { 3 } f ( x ) d x = - 5$ Then $\int _ { 1 } ^ { 8 } f ( x ) d x$ is

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The bounds m and M used in the Bounds on an Integral Theorem for $\int _ { - 1 } ^ { 2 } \left[ 2 - ( x - 1 ) ^ { 2 } \right] d x$ are

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The derivative $\frac { d } { d x } \left[ \int _ { x ^ { 2 } + 1 } ^ { 4 } \ln \left( \frac { 1 } { t } \right) d t \right]$ is

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The antiderivative $\int \frac { 1 } { x \sqrt [ 3 ] { x } } d x$ is

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The average value of $f ( x ) = | 2 x - 3 |$ on [0, 3] is

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A

Suppose S₆is the upper sum of the area enclosed by the graph $f ( x ) = 10 - x ^ { 2 },$ the x-axis, and the lines x = 0 and x = 3 by partitioning [0, 3] into six subintervals [0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2], [2, 2.5], and [2.5, 3]. Then S₆ is

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