042 Change of Variables: ⌡⌠ x 2x - 1 dx u = 2x - 1 - Cc

042
Change of Variables:
⌠
⌡ x 2x - 1 dx
u = 2x - 1 → x =
1
u +1
→ dx = 2 du
2
u + 1 1/2 1
⌠
⌡  2  u • 2 du
1
3/2
1/2
= 4⌠
⌡ (u + u ) du
1 2
2
= 4 5 u5/2 + 3 u3/2 + C
1
1
= 10 u5/2 + 6 u3/2 + C
1
1
= 10 (2x - 1)5/2 + 6 (2x - 1)3/2 + C
5
x
⌠
⌡ 2x - 1 dx
1
1
1
u = 2x - 1 → x = 2(u + 1) → dx = 2 du
When x = 1, u = 1
When x = 5, u = 9
9
1 u+1 1
⌠
⌡ 2 • u1/2 • 2 • du
1
9
1
= 4⌠
⌡
u-1/2(u + 1) du
1
9
1
1/2
-1/2
= 4⌠
⌡ (u + u ) du
1
9
1 2
= 4 3 u3/2 + 2u1/2
1
1
2
= 4 (18 + 6) - 3 + 2
1
2
= 4 24 - 2 - 3
1
64
=4 • 3
16
= 3
Alternate Choice
5
x
⌠
⌡ 2x - 1 dx
1
1
u = 2x - 1 → u2 = 2x - 1 → x = 2 (u2 + 1) → dx = u du
When x = 1, u = 1
When x = 5, u = 3
3
-1 1 2
⌠
⌡ u • 2 (u + 1) u du
1
3
1
2
= 2⌠
⌡ (u
+ 1) du
1
1 1
= 2 3 u3
3
+ u
1
1
1
= 2(9 + 3) - (3
1
1
= 2 11 - 3
1
32
=2 • 3
16
= 3
+ 1)
AP Calculus BC 1 Assignment 042 Monday, November 24, 2014 Hour
1.
∫x
x + 2 dx
2.
∫x
2 x + 1 dx
3.
∫x
4.
∫ (x + 1)
2
1 − x dx
2 − x dx
Name
5.
∫
4
6.
∫
0
2
7.
∫
0
x2 −1
dx
2x − 1
1
dx
2x + 1
x
1 + 2x
dx
2
2
8.
∫ (x − 1)
1
2 − x dx
AP Calculus BC 1 Assignment 042 Monday, November 24, 2014 Hour
1.
∫x
Ewell-Key
x + 2 dx
u=x+2 x=u-2
∫ (u − 2)
2.
Name
∫x
dx = du
∫ (u
u du =
3/ 2
)
2
4
2
4
− 2u 1 / 2 du = 5 u5/2 - 3 u3/2 + C = 5 (x + 2)5/2 - 3 (x + 2)3/2 + C
2 x + 1 dx
1
1
1
x = (2 u - 2 )
dx = 2 du
1 5/2 1 3/2
1
1
 1 3 / 2 1 1/ 2 
1
=
=
u -6 u +C
u
−
u
⋅
du
u
−
u
du




10
∫ 2 2 2
∫ 4
4

u = 2x + 1
1
1
= 10 (2x + 1)5/2 - 6 (2x + 1)3/2 + C
3.
∫x
2
1 − x dx
u=1-x
x=1-u
∫ − (1 − 2u + u
2
2
)u
1/ 2
dx = -du
du =
∫ (− u
4
5/ 2
)
2
4
2
+ 2u 3 / 2 − u 1/ 2 du = -7 u7/2 + 5 u5/2 - 3 u3/2 + C
2
= -7 (1 - x)7/2 + 5 (1 - x)5/2 - 3 (1 - x)3/2 + C
4.
∫ (x + 1)
u=2-x
2 − x dx
x=2-u
∫ − (3 − u )u
1/ 2
du =
dx = -du
∫ (u
3/ 2
)
2
2
− 3u 1 / 2 du = 5 u5/2 - 2u3/2 + C = 5 (2 - x)5/2 - 2(2 - x)3/2 + C
5.
x2 −1
∫
dx
2x − 1
u2 +1
u 4 + 2u 2 + 1
u 4 + 2u 2 − 3
→ dx = u du → x2 =
→ x2 – 1 =
2
4
4
11
2

u 4 + 2u 2 − 3 du =  u 5 + u 3 − 3u  + C
45
3

2 x − 1 → u2 = 2x – 1 → x =
u=
1 u 4 + 2u 2 − 3
1
u du = ∫
∫4⋅
u
4
(
)
 (2 x − 1)2 2(2 x − 1) 
1
1 1
2
5/ 2
3/ 2
1/ 2 
(
)
2
x
−
1
+
− 3 + C
=
2
x
−
1
+
(
2
x
−
1
)
−
3
(
2
x
−
1
)
+
C


4
5
3
4  5
3



=
or
u +1
1
u 2 + 2u + 1
u 2 + 2u − 3
2
2
→ dx = du → x =
→x –1=
u = 2x – 1 → 2x = u + 1 → x =
2
2
4
4
1
1
1
1 2
4

u 2 + 2u − 3 u −1 / 2 ⋅ du = ∫ u 3 / 2 + 2u 1 / 2 − 3u −1 / 2 du =  u 5 / 2 + u 3 / 2 − 6u 1 / 2  + C
∫
4
2
8
8 5
3

(
)
(
)
 (2 x − 1)2 2(2 x − 1) 
1
1 2
4
5/ 2
3/ 2
1/ 2 
2x − 1 
+
− 3 + C
=  (2 x − 1) + (2 x − 1) − 6(2 x − 1)  + C =
4
5
3
8 5
3



4
6.
1
∫
dx
2x + 1
0
1
u = 2x + 1 du = 2 dx
4
1
⌡(2x
2⌠
n = -2
1
+ 1)-1/2 2dx = 2 • 2(2x + 1)1/2
0
2
7.
x
∫
|
4
=3-1= 2
0
dx
1 + 2x2
0
u = 1 + 2x2
du = 4x dx
2
1
⌡(1
4⌠
1
+ 2x2)-1/2 4x dx = 2 (1 + 2x2) 1/2
0
|
2
0
3
1
=2 -2 = 1
2
8.
∫ (x − 1)
2 − x dx
1
.
u=2-x
0
.
x=2-u
0
dx = -du When x = 1, u = 1. When x = 2, u = 0
|
2
2
⌠(u - 1)u1/2 du = ⌡
⌠(u3/2 - u1/2) du = 5 u5/2 - 3 u3/2
⌡
1
1
0
1
4
= 15

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