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Calculus Chapter 6 Exam Review Find the antiderivative of the
following.
1.
€
ex +7x∫ dx 2.
€
8x5 + 6x3
+1x3∫ dx 3.
€
cos x( ) + sec2 x( )∫ dx
4.
€
xπ − π x∫ dx 5.
€
3
x x2 −1∫ dx 6.
€
5x3 + 6x2 − 8x + 9x2∫ dx
7.
€
sinh x( ) + csc2 x( )∫ dx 8.
€
10
1− x2∫ dx 9.
€
x − 2( )2∫ dx Evaluate the following definite integrals. Write
an exact answer.
10.
€
41+ x2
0
π4∫ dx 11.
€
x4 − 3x
dx1
3∫ 12.
€
2ex + 30
2∫ dx
14. Find the area of the region bounded by
€
f x( ) = x3 +1, x = 2and the x-axis. 15. A car traveling 75
miles per hour slams on its brakes and stops in 4 seconds. What is
the car’s constant
acceleration, in ft/sec2? How far does the car travel in these 4
seconds? 16. A small aircraft takes off after going down a runway
1200 feet. If the aircraft has constant acceleration
and takes 15 seconds to lift off, find its speed, in ft/sec, at
take off. 17. Use the graph at right for the
following, where x is seconds and f (x) is ft/sec.
(a) What is
€
f x( )0
20∫ dx?
(b) Consider f (x) to be the
velocity of some particle with position function s (x), what is
the particle's average velocity?
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19. Calculate the following derivatives
(a)
€
ddx
ln t5 − 3( )3x∫ dt (b)
€
ddx
cos t2 +1( )2
x3
∫ dt (c)
€
ddx
et+43
sin x∫ dt
20. Use the graph of F ' (x) = f (x) at the right and the fact
that
€
F 0( ) =1.
(a) Find F (1), F (2), F (3) and F (4). (b) Sketch a graph of y
= F (x).
21. Find the exact value for c if the area between the graph
of
€
y = x2 − c2 and the x-axis is 36. 22. A pumpkin is fired with a
slingshot with an initial velocity of 45 feet/second from an
initial height of 5
feet. Find:
(a) Time when pumpkin is at it highest point. (b) Highest value.
(c) Time when pumpkin hits the ground. (d) Velocity on impact with
the ground.
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Calculus Chapter 6 Exam Review Answers
1.
€
ex + 7ln x +C 2.
€
43x6 + 9x
23 −
x−2
2+ C 3.
€
sin x( ) + tan x( ) +C
4.
€
xπ+1
π +1−π x
lnπ+C 5.
€
3sec−1 x( ) +C 6.
€
5x2
2+ 6x − 8ln x − 9
x+ C
7.
€
cosh x( ) − cot x( ) +C 8.
€
10arcsin x( ) +C 9.
€
x3
3− 2x2 + 4x +C
10.
€
4arctan x( )0
π4
= 4arctan π4( ) − 4arctan 0( ) = 4arctan π4( )
11.
€
x3 −3x
dx1
3∫ = x
4
4− 3ln x
1
3
=3( )4
4− 3ln 3 − 1( )
4
4− 3ln1
$
% & &
'
( ) ) =
814− 3ln 3 − 14 − 0( ) = 20 − 3ln3
12.
€
2ex + 3x0
2
= 2e2 + 3 2( ) − 2e0 + 3 0( )( ) = 2e2 + 6 − 2 = 2e2 + 4
14.
€
x3 +1−1
2∫ dx = x
4
4+ x
−1
2
=24
4+ 2
$
% & &
'
( ) ) −
−1( )4
4+ −1( )
$
% & &
'
( ) ) =
274
15. 75 mph = 110 ft/sec. Acceleration = slope =
€
−1104
= −27.5 ft /sec2 Distance =
€
124( ) 110( ) = 220 feet
16. 1200 =
€
12 15( ) velocity( ) , velocity = 160 ft/sec
17a)
€
f x( )0
20∫ dx = f x( )0
4∫ dx + f x( )4
9∫ dx + f x( )9
14∫ dx + f x( )14
20∫ dx
=
€
123+ 6( )4 + 1
25( ) 6( ) + 1
25( ) −3( ) + 1
26( ) 2( )=
€
18 +15 − 7.5 + 6 = 31.5 feet.
17b) Average velocity
€
=31.520
=1.575 ft/sec.
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19a)
€
ln x5 − 3( )
19b)
€
cos x3( )2 +1" # $ %
& ' ddx
x3( ) = 3x2 cos x6 +1( )
19c)
€
esin x( )+4ddx
sin x( ) = cos x( )esin x( )+4
20a)
€
F 1( ) = F 0( ) + f x( ) dx0
1∫ =1+ 1( ) 2( ) = 3 b)
€
F 2( ) = F 1( ) + f x( ) dx1
2∫ = 3+ 12 1( ) 2( ) = 4
€
F 3( ) = F 2( ) + f x( ) dx2
3∫ = 4 + 12 1( ) −1( ) = 3.5
€
F 4( ) = F 3( ) + f x( ) dx3
4∫ = 3.5 + 1( ) −1( ) = 2.5
21. The graph of
€
y = x2 − c2 is a parabola below the x-axis and the
zeros of
€
y = x2 − c2 are
€
x = ±c so
€
x2 − c2−c
c∫ dx = −36 . Since
€
y = x2 − c2 is a parabola symmetrical to the y-axis, then
€
y = x2 − c2 is an even function and can rewrite the integral
as
€
2 x2 − c20
c∫ dx = −36
€
⇒ x2 − c20
c∫ dx = −18
€
⇒x3
3− c2x
0
c
= −18
€
⇒c3
3− c2 c( ) = −18
€
⇒c3
3− c3 = −18
€
⇒−2c3
3= −18
€
⇒ c3 = 27 , so c = 3.
22.
€
a t( ) = −32 ft /sec2 ,
€
v t( ) = −32t + 45 ft /sec ,
€
h t( ) = −16t2 + 45t + 5 feet 22a) Highest point occurs with v
(t) = 0, giving
€
−32t + 45 = 0 when
€
t = 4532 seconds.
22b) Highest value occurs at
€
h 4532( ) = −16 4532( )2
+ 45 4532( ) + 5 ≈ 36.641 feet 22c) Hits the ground when h (t) =
0, when t = 2.92 seconds (using the calculator). 22d) Velocity on
impact is
€
v 2.92( ) = −32 2.92( ) + 45 = −48.425 ft/sec.
