COLLEGE ROUND ONE⇒ You will have two minutes to evaluate each of the fifteen definiteintegrals that will displayed one at a time on this screen. All answersmust be simplified. At the end of the two minutes, all hands must goup and judges will grade your answers immediately. For each correctanswer, you will receive one raffle ticket to be entered for prizes thatwill be drawn after dinner.

At most five participants will move to the finals – to be determined bythe total number of correct answers and tiebreaking criteria ifnecessary. Everyone moving to the finals will receive $25.

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INTEGRAL #1

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #1∫ 1210

3

√(1

2x− 5

)4

dx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #1∫ 1210

3

√(1

2x− 5

)4

dx

=

∫ 1210

(1

2x− 5

)4/3

dx

= 2

∫ 10

u4/3 du[u =

1

2x− 5, du =

1

2dx

]

= 2

[3u7/3

7

]10

=6

7

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #2

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #2∫π/20

(3x+ 2) sinxdx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #2∫π/20

(3x+ 2) sinxdx[

integrate by parts:u = 3x+ 2

du = 3dx,

dv = sin xdxv = − cos x

]

=[− (3x+ 2) cos x

]π/20

+ 3

∫π/20

cos xdx

=[− (3x+ 2) cos x

]π/20

+ 3[

sin x]π/20

= 5

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #3

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #3∫ 20

ex · e2x · e3x · e4x dx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #3∫ 20

ex · e2x · e3x · e4x dx

=

∫ 20

e10x dx

=

[e10x10

]20

=e20 − 1

10

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #4

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #4∫√3

0

x3√x2 + 1

dx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #4∫√3

0

x3√x2 + 1

dx

=1

2

∫ 41

u− 1√u

du[u = x2 + 1, x2 = u− 1, 2xdx = du

]=

1

2

∫ 41

(u1/2–u−1/2

)du

=1

2

[2u3/2

3− 2u1/2

]41

=4

3

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #5

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #5∫ 10

(√x+ 2

)(5x+ 3

)dx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #5∫ 10

(√x+ 2

)(5x+ 3

)dx

=

∫ 10

(5x3/2 + 3x1/2 + 10x+ 6

)dx

=[2x5/2 + 2x3/2 + 5x2 + 6x

]10

= 15

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #6

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #6∫π/20

√sinx− sin3 xdx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #6∫π/20

√sinx− sin3 xdx

=

∫π/20

√sin x(1− sin2 x)dx =

∫π/20

√sin x · cos2 xdx

=

∫π/20

√sin x · cos xdx =

∫ 10

√udu [ u = sin x ]

=

[2u3/2

3

]10

=2

3

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #7

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #7∫ 10

x2 + 3x+ 3

x+ 1dx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #7∫ 10

x2 + 3x+ 3

x+ 1dx

=

∫ 10

(x+ 2+

1

x+ 1

)dx [ long division ]

=

[x2

2+ 2x+ ln(x+ 1)

]10

=5

2+ ln 2 or 5+ 2 ln 2

2or 5+ ln 4

2

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #8

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #8∫π0

sinx · sin x

2dx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #8∫π0

sinx · sin x

2dx

=

∫π0

(2 sin x

2cos x

2

)· sin x

2dx [ sin 2θ = 2 sin θ cos θ ]

= 2

∫π0

sin2 x

2cos x

2dx

[u = sin x

2, du =

1

2cos x

2dx

]

= 4

∫ 10

u2 du = 4

[u3

3

]10

=4

3

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #9

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #9∫ 10

xπ · πe · xe · eπ dx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #9∫ 10

xπ · πe · xe · eπ dx

= πe · eπ∫ 10

xπ+e dx

= πe · eπ[

xπ+e+1

π+ e + 1

]10

=πe · eπ

π+ e + 1

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #10

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #10∫π/30

(sinx+ tanx)(cosx+ sec2 x)dx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #10∫π/30

(sinx+ tanx)(cosx+ sec2 x)dx[u = sin x+ tan x, du = (cos x+ sec2 x)dx

]=

∫ 3√3/2

0

udu

=

[u2

2

]3√3/2

0

=27

8

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #11

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #11∫ 41

1

2√x√2+

√x

dx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #11∫ 41

1

2√x√2+

√x

dx

=

∫ 43

1√u

du[u = 2+

√x, du =

1

2√x

du]

=[2√u]43

= 4− 2√3

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #12

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #12∫π/30

sec4 x tanxdx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #12∫π/30

sec4 x tanxdx

=

∫π/30

sec3 x · sec x tan xdx

=

∫ 21

u3 du [ u = sec x, du = sec x tan xdx ]

=

[u4

4

]21

=15

4

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #13

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #13∫ 21

xe2x + lnx

xdx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #13∫ 21

xe2x + lnx

xdx

=

∫ 21

(xe2xx

+ln x

x

)dx

=

∫ 21

(e2x + ln x

x

)dx

=

[e2x2

+(ln x)2

2

]21

=e4 − e2 + (ln 2)2

2

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #14

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #14∫ 10

arctanxdx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #14∫ 10

arctanxdx[

integrate by parts:u = arctan x

du = 1

x2+1dx

,dv = dxv = x

]

=[x arctan x

]10−

∫ 10

x

x2 + 1dx =

π

4−

[1

2ln

(x2 + 1

)]10

=π

4−

ln 2

2or π− 2 ln 2

4or π− ln 4

4

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #15

READY,GET SET,…

2:002 0 1 4 U of S I N T E G R A T I O N B E E

U of S I N

INTEGRAL #15∫ 71

2014

2020x + 14

x

+14

20x + 14

x

dx

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N

INTEGRAL #15∫ 71

2014

2020x + 14

x

+14

20x + 14

x

dx

=

∫ 71

2014

xdx [ Simplify! ]

= 2014[

ln x]71

= 2014 ln 7

2 0 1 4 U of S I N T E G R A T I O N B E EU of S I N