y ysx y4 yt yu yx yx yx

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y ysx y4 yt yu yx yx yx
SECTION 5.4
5.4
THE FUNDAMENTAL THEOREM OF CALCULUS
A Click here for answers.
S Click here for solutions.
1. Sketch the area represented by
x
tx y 2 cos t dt
Then find tx in two ways: (a) by using Part 1 of the Fundamental Theorem and (b) by evaluating the integral using Part 2
and then differentiating.
■ Use Part 1 of the Fundamental Theorem of Calculus to find
the derivative of the function.
2–12
Copyright © 2013, Cengage Learning. All rights reserved.
3. tx y
1
dt
1 t4
5. tt y
7. hx y
4. tu y
6. Fx y (2 su )
8. hx y
y
12. y y
■
x2
u
4
8
x
sx
du
s2
ds
s 1
sin t
dt
t
1
t
1x
2
y
17
11. y y
5x1
tan x
st 3 1 dt
sinx 2 dx
0
9. y 2
1
sin x
5
x
t 2 1 20 dt
y
10. y ■
x
2. tx 1
sin 4 t dt
sint 4 dt
0
1
du
u2 5
t cost 3 dt
■
■
THE FUNDAMENTAL THEOREM OF CALCULUS
■
■
■
■
■
■
■
■
1
2
■
SECTION 5.4
THE FUNDAMENTAL THEOREM OF CALCULUS
5.4
ANSWERS
E Click here for exercises.
(a), (b) 2 + cos x
1.
2.
20
g (x) = x2 − 1
4.
g (u) =
6.
F (x) = − (2 +
1
1 + u4
√ 8
x)
√
x
2 (x + 1)
2 sin x2
dy
10.
=−
dx
x
8.
Copyright © 2013, Cengage Learning. All rights reserved.
12.
S Click here for solutions.
h (x) =
√
x3 + 1
3. g
(x) =
5. g
(t) = sin t2
− sin4 (1/x)
x2
dy
= − sin tan4 x sec2 x
9.
dx
7. h
11.
dy
= sin x cos x cos sin3 x
dx
(x) =
5
dy
=
dx
25x2 + 10x − 4
SECTION 5.4
5.4
THE FUNDAMENTAL THEOREM OF CALCULUS
SOLUTIONS
E Click here for exercises.
10.
1.
Let u = x2 . Then
d
dy
=
dx
dx
(b) g (x) =
x
⇒ g (x) = 2 + cos x
π
(2 + cos t) dt
π
(2 + cos t) dt = [2t + sin t]xπ
x
= (2x + sin x) − (2π + 0) = 2x + sin x − 2π
2.
3.
4.
5.
6.
7.
so g (x) = 2 + cos x.
x
20
20
g (x) = 1 t2 − 1
dt ⇒ g (x) = x2 − 1
x √
√
g (x) = −1
t3 + 1 dt ⇒ g (x) = x3 + 1
u
1
1
g (u) =
dt ⇒ g (u) =
4
1
+
t
1
+
u4
π
t
2
g (t) = 0 sin x dx ⇒ g (t) = sin t2
4
x
√ 8
√ 8
F (x) = x (2 + u) du = − 4 (2 + u) du ⇒
√ 8
F (x) = − (2 + x)
du
1
1
. Then
= − 2 , so
x
dx
x
1/x
u
d
d
du
sin4 t dt =
sin4 t dt ·
dx 2
du 2
dx
Let u =
Copyright © 2013, Cengage Learning. All rights reserved.
= sin4 u
− sin4 (1/x)
du
=
dx
x2
√
1
du
= √ , so
x. Then
dx
2 x
√x
d
s2
ds
h (x) =
2
dx 1
s +1
u
du
u2 du
d
s2
ds
·
=
=
du 1 s2 + 1
dx
u2 + 1 dx
√
1
x
x
√ =
=
x+12 x
2 (x + 1)
8.
Let u =
9.
Let u = tan x. Then
du
= sec2 x, so
dx
17
tan x
d
d
sin t4 dt = −
sin t4 dt
dx tan x
dx 17
u
d
du
= −
sin t4 dt ·
du 17
dx
du
= − sin u4
dx
= − sin tan4 x sec2 x
π
d
sin t
dt = −
t
dx
x2
x2
π
sin t
dt
t
u
du
sin u du
sin t
dt ·
=−
·
t
dx
u
dx
2
2
2 sin x
sin x
· 2x = −
= −
x2
x
dt
11. Let t = 5x + 1. Then
= 5, so
dx
5x+1
d t 1
dt
1
d
du
=
du ·
dx 0
u2 − 5
dt 0 u2 − 5
dx
= −
(a) g (x) =
du
= 2x, so
dx
d
du
π
1 dt
t2 − 5 dx
5
=
25x2 + 10x − 4
=
12.
du
= cos x, so
dx
u
dy du
d
du
dy
=
=
t cos t3 dt ·
dx
du dx
du −5
dx
Let u = sin x. Then
du
= sin x cos sin3 x cos x
= u cos u3
dx
= sin x cos x cos sin3 x
■
3

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