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 attempted 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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I N T E G R A T I O N
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INTEGRAL #1
READY,
GET SET,…
2 : 30
2 0 1 2
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I N T E G R A T I O N
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INTEGRAL #1
∫
(
1 x2 + 1
x3
x2
)9
dx
2 : 30
2 0 1 2
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I N T E G R A T I O N
B E E
INTEGRAL #1
∫
(
1 x2 + 1
x3
x2
)9
dx
[
(
)9
1
1
1
=
dx
u = 1 + 2,
1+ 2
x3
x
x
∫
1
1 u10
9
=−
u du = − ·
+C
2
2 10
∫
2
du = − 3
x
]
)10
(
1
1
1+ 2
= −
20
x
2 0 1 2
U
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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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I N T E G R A T I O N
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INTEGRAL #2
∫
sin x
dx
4
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 #2
∫
sin x
dx
4
cos x
∫
1
du [ u = cos x,
=−
u4
du = − sin x ]
1
= 3 +C
3u
=
1
3 cos3 x
2 0 1 2
U
+C
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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I N T E G R A T I O N
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INTEGRAL #3
∫
4
dx
2
4x + 4x + 1
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
∫
4
dx
2
4x + 4x + 1
∫
=
∫
=
4
dx [ u = 2x + 1,
(2x + 1)2
du = 2 dx ]
2
du
u2
2
2
=− +C= −
+C
u
2x + 1
2 0 1 2
U
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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
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INTEGRAL #4
∫
ex + 2e2x + 3e3x
dx
4x
4e
2 : 30
2 0 1 2
U
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I N T E G R A T I O N
B E E
INTEGRAL #4
∫
ex + 2e2x + 3e3x
dx
4x
4e
∫(
=
∫(
=
ex
2e2x 3e3x
+
+
4e4x 4e4x 4e4x
)
dx
1 −3x 1 −2x 3 −x
e
+ e + e
4
2
4
e−3x
)
dx
3e−x
= −
−
−
+C
12
4
4
2 0 1 2
U
e−2x
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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INTEGRAL #5
∫ √
√
3 1+ x
√
dx
x
2 : 30
2 0 1 2
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I N T E G R A T I O N
B E E
INTEGRAL #5
∫ √
√
3 1+ x
√
dx
x
∫
=6
√
[
u=1+
u du
√
x,
]
1
du = √ dx
2 x
= 4u3/2 + C
= 4(1 +
2 0 1 2
√
U
x)3/2 + C
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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INTEGRAL #6
∫
ex sin ex cos ex 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 #6
∫
ex sin ex cos ex dx
∫
=
u du [ u = sin ex,
du = ex cos ex dx ]
sin2 ex
u2
=
+C=
+ C or
2
2
2 0 1 2
U
of
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−
cos2 ex
2
+C
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
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INTEGRAL #7
∫
cot x − sec x
dx
cot x
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
∫
cot x − sec x
dx
cot x
∫(
=
cot x sec x )
−
dx
cot x cot x
∫
=
(1 − sec x tan x) dx
= x − sec x + C
2 0 1 2
U
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I N T E G R A T I O N
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INTEGRAL #8
READY,
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2 : 30
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INTEGRAL #8
∫
(x + 1)3 + (x + 2)3
dx
3
3
(x + 1) (x + 2)
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 #8
∫
(x + 1)3 + (x + 2)3
dx
3
3
(x + 1) (x + 2)
∫(
3
3
(x + 1)
(x + 2)
+
(x + 1)3(x + 2)3 (x + 1)3(x + 2)3
)
∫(
1
1
=
+
dx
3
3
(x + 2)
(x + 1)
=
)
dx
1
1
= −
−
+C
2
2
2(x + 2)
2(x + 1)
2 0 1 2
U
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I N T E G R A T I O N
B E E
INTEGRAL #9
READY,
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2 : 30
2 0 1 2
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INTEGRAL #9
∫ √
tan x
dx
2
1 − sin x
2 : 30
2 0 1 2
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I N T E G R A T I O N
B E E
INTEGRAL #9
∫ √
tan x
dx
2
1 − sin x
∫√
tan x
dx =
2
cos x
=
∫
=
√
∫
√
tan x sec2 x dx
[
u du u = tan x,
du = sec2 x dx
]
2u3/2
2 tan3/2 x
=
+C=
+C
3
3
2 0 1 2
U
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I N T E G R A T I O N
B E E
INTEGRAL #10
READY,
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2 : 30
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INTEGRAL #10
∫
√
3
x
√
dx
√
3
x 33 + 3 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 #10
∫
√
3
x
√
dx
√
3
x 33 + 3 x
∫
x−2/3
= √
dx
√
3
3
33 + x
[
∫
√
3
√
= 3 du
u = 33 + 3 x,
u
2/3
9u
=
2
2 0 1 2
1 −2/3
du = x
dx
3
]
√
3
9(33 + x)2/3
+C=
+C
2
U
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