dx - Kirkwood School District

AP C ALCULUS
T HE F UNDAMENTAL T HEOREMS A
Z
1
1.
0
¯1
¯
(2x + 1) dx = (x2 − x)¯¯ = (1 − 1) − (0) = 0
0
Z
1
2.
Z
(x4 + x)(4x3 + 1) dx
1
=
(4x7 + 5x4 + x) dx
0
0
µ
=
µ
=
=
¶ ¯1
¯
1 8
1
x + x5 + x2 ¯¯
2
2
0
1
1
+1+
2
2
¶
−0
2
3. u = x + 1 −→ du = dx
x=u+1
x = 1 −→ u = 0
x = 2 −→ u = 1
Z
2
Z
√
x x − 1 dx
1
=
1
u1/2 (u + 1) du
0
Z
1
=
³
´
u3/2 + u1/2 du
0
µ
=
µ
=
Z
1
cos πx dx
=
0
¶
−0
¯1
¯
1
sin πx¯¯
π
0
µ
=
=
2 2
+
5 3
16
15
=
4.
¶ ¯1
2 5/2 2 3/2 ¯¯
u + u
¯
3
3
0
¶ µ
¶
1
1
sin π −
sin 0
π
π
0
1
5. u = 1 +
1
1
−→ du = − 2 dx
x
x
x = 1 −→ u = 2
5
x = 4 −→ u =
4
Z 4 r
1
1
1 + dx
2
x
1 x
Z
=
5/4
−
u1/2 du
2
Z
2
=
u1/2 du
5/4
Z
π/3
6.
0
=
¯2
2 3/2 ¯¯
u ¯
3
5/4
=
√
√
4 2 5 5
−
3
12
Z
sin x
dx
cos2 x
π/3
=
tan x sec x dx
0
=
¯π/3
¯
sec x¯¯
0
=
7. u = 1 + 2x −→
1
1
du = dx
2
x = 0 −→ u = 1
x = 13 −→ u = 27
Z
0
13
p
3
dx
(1 + 2x)2
dx
=
1
2
Z
27
u−2/3 du
1
=
¯27
3 1/3 ¯¯
u ¯
2
1
=
9 3
−
2 2
=
3
2
8. u = x − 2 −→ du = dx
x = 0 −→ u = −2
x = 4 −→ u = 2
Z
4
0
Z
dx
dx
(x − 2)3
2
=
u−3 du
−2
=
¯2
¯
1
− u−2 ¯¯
2
−2
=
¯2
1 ¯¯
− 2¯
2u −2
=
µ
¶ µ
¶
1
1
−
− −
8
8
=
0
1
du = dx
2
9. u = x2 + a2 −→
x = 0 −→ u = a2
x = a −→ u = 2a2
Z
a
p
x x2 + a2 dx
1
2
=
0
Z
2a2
u1/2 du
a2
=
¯2a2
1 2 3/2 ¯¯
u ¯
23
a2
=
1√ 6 1√ 6
8a −
a
3
3
10. It is critical that you are able to separate absolute valued integrals but whenever we have the absolute value of a linear
expression, the preferred method is to draw the graph and add up the triangles.
Solution by separating the integral:
Z
Z
5
|x − 3| dx
−2
Z
3
=
5
(3 − x) dx +
−2
(x − 3) dx
3
=
µ
µ
¶ ¯3
¶ ¯5
¯
1 2
1 2 ¯¯
3x − x ¯ +
x − 3x ¯¯
2
2
−2
3
=
µ
¶
µ
¶ µ
¶
9
25
9
9−
− (−6 − 2) +
− 15 −
−9
2
2
2
=
29
Z2
5
Solution via triangles:
|x − 3| dx
=
1
1
(5)(5) + (2)(2)
2
2
=
29
2
−2
3
11. First, divide the numerator by the denominator using polynomial long division or synthetic division.
¶
Z 1 3
Z 1µ
x −1
2
2
dx =
x −x+1−
dx
x+1
0 x+1
0
·
=
¸ ¯1
¯
1 2 1 2
x − x + x − 2 ln |x + 1| ¯¯
3
2
0
µ
=
=
Z
5
− 2 ln 2
6
=
¯π/2
¯
1
− cos 2x¯¯
2
0
=
µ
¶ µ
¶
1
1
− cos π − − cos 0
2
2
=
1
π/2
12.
sin 2x dx
0
13. u = 3x2 − 1 −→
¶
1 1
− + 1 − 2 ln 2 − (0 − 0 + 0 − 2 ln 1)
3 2
1
du = x dx
6
x = 1 −→ u = 2
x = 3 −→ u = 26
Z
3
1
x
dx
(3x2 − 1)3
=
1
6
Z
26
u−3 du
2
=
¯26
1 u−2 ¯¯
6 −2 ¯2
=
¯26
1 ¯¯
−
12u2 ¯2
=
7
338
14. u = x + 3 −→ du = dx
x = u − 3 −→ x + 2 = u − 1
x = −2 −→ u = 1
x = 1 −→ u = 4
Z
1
√
(x + 2) x + 3 dx
Z
4
=
−2
u1/2 (u − 1) du
1
Z
4
³
=
´
u3/2 − u1/2 du
1
µ
=
·
=
=
¶ ¯4
2 5/2 2 3/2 ¯¯
u − u
¯
5
3
1
¸ ·
¸
2
2
2 2
(32) − (8) −
−
5
3
5 3
116
5
4
Z
15.
2
4
w4 − w
dw
w3
Z
4
=
(w − w−2 ) dw
2
µ
=
¶ ¯4
¯
1 2
w + w−1 ¯¯
2
2
=
µ
¶ µ
¶
1
1
8+
− 2+
4
2
=
23
4
5