COLLEGE ROUND ONE
⇒
You will have two minutes to evaluate each of the fifteen
definite integrals that will displayed one at a time on this screen.
All answers must be simplified. At the end of the two minutes,
all hands must go up and judges will grade your answers
immediately. For each correct answer, you will receive one raffle
ticket to be entered for prizes that will be drawn after dinner.
At most five participants will move to the finals—to be
determined by the total number of correct answers and
tiebreaking criteria if necessary. Everyone moving to the finals
will receive $25.
2 0 1 3
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INTEGRAL #1
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2 0 1 3
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INTEGRAL #1
∫ 1 (√
3
)
√
4
x4 + x3 dx
0
2 : 00
2 0 1 3
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INTEGRAL #1
∫ 1 (√
3
0
)
√
4
x4 + x3 dx
∫1 (
=
)
x4/3 + x3/4 dx
0
[
3x7/3 4x7/4
=
+
7
7
]1
0
= 1
2 0 1 3
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INTEGRAL #2
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INTEGRAL #2
∫ π/2
(
sin x + cos x
)2
dx
0
2 : 00
2 0 1 3
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INTEGRAL #2
∫ π/2
0
(
sin x + cos x
∫ π/2
=
(
)2
dx
)
sin x + 2 sin x cos x + cos x dx
2
2
0
∫ π/2
=
[
]π/2
(1 + 2 sin x cos x) dx = x + sin x
0
0
=
2
π+2
π
+1=
2
2
2 0 1 3
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INTEGRAL #3
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2 : 00
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INTEGRAL #3
∫1
)3
( 4
x x + 1 dx
3
0
2 : 00
2 0 1 3
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INTEGRAL #3
∫1
)3
( 4
x x + 1 dx
3
0
∫2
[
1
u3 du
=
4 1
[ 4 ]2
u
=
16 1
4
u = x + 1,
3
du = 4x dx
]
15
=
16
2 0 1 3
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INTEGRAL #4
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INTEGRAL #4
∫e
1
(
1
(
)2) dx
x 1 + ln x
2 : 00
2 0 1 3
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INTEGRAL #4
∫e
1
(
1
(
)2) dx
x 1 + ln x
∫1
=
0
[
1
1 + u2
u = ln x,
[
]1
= arctan u
=
1
du = dx
x
]
0
π
4
2 0 1 3
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INTEGRAL #5
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2 : 00
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INTEGRAL #5
∫4
3
2x
dx
(x + 2)(x − 2)
2 : 00
2 0 1 3
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INTEGRAL #5
∫4
3
2x
dx
(x + 2)(x − 2)
∫4 (
=
3
1
1
+
x+2 x−2
)
dx
[
]4
= ln(x + 2) + ln(x − 2)
= ln 6 + ln 2 − ln 5 = ln
2 0 1 3
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3
12
5
I N T E G R A T I O N
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INTEGRAL #6
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2 : 00
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INTEGRAL #6
∫ ln 2
xex dx
0
2 : 00
2 0 1 3
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INTEGRAL #6
∫ ln 2
xex dx
0
[
integrate by parts:
u = x
du = dx
,
dv = ex dx
]
v = ex
[ ]ln 2 ∫ ln 2
= xex
−
ex dx
0
0
[
]ln 2
x
x
= xe − e
= 2 ln 2 − 1 = ln 4 − 1
0
2 0 1 3
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INTEGRAL #7
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INTEGRAL #7
∫2
1
2x3 + 3x4 + 4x5
dx
2
x
2 : 00
2 0 1 3
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INTEGRAL #7
∫2
1
2x3 + 3x4 + 4x5
dx
2
x
∫2
=
(
2
3
)
2x + 3x + 4x dx
1
[
]2
2
3
4
= x +x +x
1
= 25
2 0 1 3
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INTEGRAL #8
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INTEGRAL #8
∫ ln 2
0
e2x
dx
2x
e + 2013
2 : 00
2 0 1 3
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INTEGRAL #8
∫ ln 2
0
e2x
dx
2x
e + 2013
∫ 2017
1
du
=
du
2 2014 u
]2017
[
ln u
=
2 2014
=
[
ln 2017 − ln 2014
2 0 1 3
2
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u = e2x + 2013,
1 2017
= ln
= ln
2 2014
du = 2e2x dx
√
]
2017
2014
I N T E G R A T I O N
B E E
INTEGRAL #9
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2 : 00
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INTEGRAL #9
∫1
0
(√
1
x2 + 1
)3 dx
2 : 00
2 0 1 3
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INTEGRAL #9
∫1
0
(√
1
x2 + 1
)3 dx
[
dx = sec2 θ dθ,
x = tan θ,
∫ π/4
=
0
∫ π/4
=
0
2 0 1 3
1
· sec θ dθ =
∫ π/4
2
sec3 θ
]
√
x2 + 1 = sec θ
0
1
sec θ
dθ
√
]π/4
2
cos θ dθ = sin θ
=
0
2
[
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INTEGRAL #10
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2 : 00
2 0 1 3
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INTEGRAL #10
∫ π/4
tan2 x dx
0
2 : 00
2 0 1 3
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INTEGRAL #10
∫ π/4
tan2 x dx
0
∫ π/4
=
(
)
2
sec x − 1 dx
0
[
]π/4
= tan x − x
dx
0
π 4−π
= 1− =
4
4
2 0 1 3
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INTEGRAL #11
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2 : 00
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INTEGRAL #11
∫1 (
)
(
)2
(
)
2
x − 2 + x2 + x + 2
dx
0
2 : 00
2 0 1 3
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I N T E G R A T I O N
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INTEGRAL #11
∫1 (
0
)
(
)2
(
)
2
x − 2 + x2 + x + 2
dx
∫1
=
(
)
3x + 8 dx
2
0
[
]1
3
= x + 8x
0
= 9
2 0 1 3
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INTEGRAL #12
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2 : 00
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INTEGRAL #12
∫ e2
ln x dx
1
2 : 00
2 0 1 3
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INTEGRAL #12
∫ e2
ln x dx
1
[
integrate by parts:
u = ln x
du =
[
]e2 ∫ e2
= x ln x −
dx
1
1
x
dx
,
dv = dx
]
v = x
1
[
]e2
= x ln x − x = e2 + 1
1
2 0 1 3
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INTEGRAL #13
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2 : 00
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INTEGRAL #13
∫4
1
1
) dx
√ (
√
x x+2 x+1
2 : 00
2 0 1 3
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of
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I N T E G R A T I O N
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INTEGRAL #13
∫4
1
1
) dx
√ (
√
x x+2 x+1
∫4
1
= √ (√
)2 dx
1
x x+1
[
∫3
√
1
=2
du u = x + 1,
2
2 u
1
du = √ du
2 x
]
[
2 ]3
1
= −
=
u 2
3
2 0 1 3
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INTEGRAL #14
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INTEGRAL #14
∫ π/3
sec x · ln(sec x + tan x) dx
0
2 : 00
2 0 1 3
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INTEGRAL #14
∫ π/3
sec x · ln(sec x + tan x) dx
0
∫ ln(2+√3)
=
[
u du u = ln(sec x + tan x),
du = sec x dx
]
0
√
]
ln
2+
3)
u2 (
=
=
2 0
[
2 0 1 3
U
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( (
√ ) )2
ln 2 + 3
2
I N T E G R A T I O N
B E E
INTEGRAL #15
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INTEGRAL #15
∫ π/4
( x
)
x
2
e tan x + e sec x dx
0
2 : 00
2 0 1 3
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INTEGRAL #15
∫ π/4
0
( x
)
x
2
e tan x + e sec x dx
∫ π/4
=
0
[
]
d x
(e tan x) dx
product rule
dx
[
]π/4
= ex tan x
0
= eπ/4
2 0 1 3
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