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Module 2: Now You Try II
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Module 2: Now You Try II
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Module 2: Now You Try II
Compute:
1.
∫
2
1
(
x
−
1
)
d
x
Solution:
▶ Show
∫
2
1
(
x
−
1
)
d
x
=
(
x
2
2
−
x
)
|
2
1
=
(
2
2
2
−
2
)
−
(
1
2
2
−
1
)
2.
∫
π
4
0
sec
2
θ
d
θ
Solution:
▶ Show
∫
π
4
0
sec
2
θ
d
θ
=
(
tan
θ
)
|
π
4
0
=
tan
(
π
4
)
−
tan
(
0
)
=
1.
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Module 2: Definite Integrals
Module 2: Definite versus Indefinite Integral