THE HIGH SCHOOL FINALS
⇒
The Finals will be conducted in rounds. One at a time, each
remaining contestant will have two and a half minutes to
compute an indefinite integral. If answered correctly, the
contestant remains in the competition. Once every remaining
contestant has a empted one problem, a round is completed. If
during any round, all contestants are unable to complete a
problem correctly, all contestants will remain in the competition
for another round.
The last person remaining wins an additional $75 and will be
crowned the Integration Champion!
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #1
READY,
GET SET,…
2 : 30
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #1
∫
√
sin x cos x dx
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #1
∫
√
sin x cos x dx
∫
=−
√
[
u du u = cos x,
]
du = − sin x dx
2u3/2
=−
+C
3
2 cos3/2 x
+C
= −
3
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #2
READY,
GET SET,…
2 : 30
2 0 1 2
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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 #2
∫
1
√
√ 5 dx
x (2012 + x)
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #2
∫
1
√
√ 5 dx
x (2012 + x)
[
∫
√
1
=2
du u = 2012 + x,
u5
(
)
1
=2 − 4 +C
4u
1
du = √ dx
2 x
]
1
= − (
√ )4 + C
2 2012 + x
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #3
READY,
GET SET,…
2 : 30
2 0 1 2
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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 #3
∫
)2
x 2x + 1 dx
(
3
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #3
∫
)2
x 2x + 1 dx
(
3
∫
=
6
3
)
x 4x + 4x + 1 dx
∫
=
(
(
)
4x + 4x + x dx
7
4
x8 4x5 x2
=
+
+ +C
2
5
2
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #4
READY,
GET SET,…
2 : 30
2 0 1 2
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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 #4
∫
x+1
(x2 + 2x + 2012)9
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #4
∫
x+1
(x2 + 2x + 2012)9
1
=
2
∫
du
u9
[
u = x2 + 2x + 2012,
]
dx = 2(x + 1) dx
(
)
1
1
=
− 8 +C
2
8u
1
+C
= −
16(x2 + 2x + 2012)8
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #5
READY,
GET SET,…
2 : 30
2 0 1 2
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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 #5
∫
x
√
dx
7x2 + 7
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #5
∫
x
√
dx
7x2 + 7
1
=
14
∫
[
du
√
u
2
u = 7x + 7,
]
du = 14x dx
√
1
·2 u+C
=
14
√
=
2 0 1 2
7x2 + 7
+C
7
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #6
READY,
GET SET,…
2 : 30
2 0 1 2
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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 #6
∫
1
x3
√
3
1
1 + 2 dx
x
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #6
∫
1
x3
√
3
1
1 + 2 dx
x
[
1
1
=−
u1/3 du u = 1 + 2 ,
2
x
∫
du = −
2
dx
3
x
]
1 3u4/3
=− ·
+C
2
4
(
)4/3
1
3
1+ 2
+C
= −
8
x
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #7
READY,
GET SET,…
2 : 30
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #7
∫
√
sin x
√
dx
x
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #7
∫
√
sin x
√
dx
x
[
∫
= 2 sin u du
√
u = x,
1
du = √ dx
2 x
]
= −2 cos u + C
√
= −2 cos x + C
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #8
READY,
GET SET,…
2 : 30
2 0 1 2
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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 #8
∫
sin x − cos x
√
dx
sin x + cos x
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #8
∫
sin x − cos x
√
dx
sin x + cos x
∫
du
√
=−
u
[
u = sin x + cos x,
]
du = (cos x − sin x) dx
√
= −2 u + C
√
= −2 sin x + cos x + C
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #9
READY,
GET SET,…
2 : 30
2 0 1 2
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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 #9
∫
x5
√
x3 + 1 dx
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #9
∫
x5
√
x3 + 1 dx
[
3
one of many possible subs: u = x + 1,
∫
√
1
1
=
(u − 1) u du =
3
3
=
1
3
2 0 1 2
(
3
2(x + 1)
5
U
of
S
5/2
∫(
u
3
−
3/2
2(x + 1)
3
−u
3/2
1/2
2
]
du = 3x dx
)
du
)
+C
I N T E G R A T I O N
B E E
INTEGRAL #10
READY,
GET SET,…
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
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INTEGRAL #10
∫
(sin x + cos x)2 dx
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #10
∫
(sin x + cos x)2 dx
∫
=
∫
=
(
)
sin x + 2 sin x cos x + cos x dx
2
2
[ ∫
]
(1 + 2 sin x cos x) dx = (1 + sin 2x) dx
= x + sin2 x + C or x − cos2 x + C or x −
2 0 1 2
U
of
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cos 2x
I N T E G R A T I O N
2
+C
B E E
INTEGRAL #11
READY,
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2 : 30
2 0 1 2
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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 #11
∫
1
(
1
√
2+√
x x
x
)4
dx
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #11
∫
1
(
1
√
2+√
x x
x
)4
dx
[
1
4
√
= −2 u du u = 2 +
,
x
∫
1
√
du = −
x x
]
−2u5
=
+C
5
(
)5
2
1
= −
2+√
+C
5
x
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #12
READY,
GET SET,…
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #12
∫
sin 2x
dx
3
cos x
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #12
∫
sin 2x
dx
3
cos x
∫
∫
sin x
1
2 sin x cos x
dx = 2
·
dx
=
cos3 x
cos x cos x
∫
= 2 sec x tan x dx
= 2 sec x + C
2 0 1 2
U
of
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I N T E G R A T I O N
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