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Unformatted text preview: 216 Section 5.3 19. Divide the shaded area as follows. 2 3 (x 2 (b) Area 6x 8) dx 0 y
2.25 [F(2) y = 3x – x 2 20
3 R1 (x 2 6x 8) dx 2 F(0)]
0 [F(3) 6 F(2)] 20
3 22
3 x R2 4 22. –4 Note that an antiderivative of 3x
3 Area of R1 x 2) dx (3x
0 4 Area of R2 32
13
x
x.
2
3
9
9
0
2
2 x 2 is F(x) F(3) F(0) [0, 2] by [ 5, 3] x 2) dx (3x 13
x
3 F(x) 3 [F(4)
8
3 F(3)]
9
2 2 (a) 11
6 9
2 Total shaded area x2 An antiderivative of
52
x
2 ( x2 5x 5x 4 is 4x.
4) dx F(2) 2
3 F(0) 11
6 0
1 19
3 (b) Area 2 ( x2 5x ( x2 4) dx 0 5x 4) dx 1 [F(1)
20. Divide the shaded area as follows. 2
3 0 F(0)] [F(2) F(1)] 0 2
3 11
6 3 x2 13
x.
3 11
6 y 23. 3
y= x2 – 2x R2 [0, 3] by [ 3, 2] x 3 R1 x 2 is F(x) An antiderivative of 2x Note that an antiderivative of x 2
2 Area of R1 (x 2 13
x
3 2x is F(x) 3 x 2. (a) 2x) dx [F(2) 2 (x 2 F(0) 0 3 x 2 ) dx (2x 0 0 (2x 4
3 2x) dx x 2) dx 2 [F(2) 4
3 0 F(3) 0 F(0)] 4
3 3 x 2 ) dx (b) Area 0 Area of R2 (2x
0 F(0)]
0 [F(3)
4
3 0 F(2)] 8
3 2 F(3)
0 24. F(2)
4
3
4
3 Total shaded area 4
3
4
3 8
3
[0, 5] by [ 5, 5] 21. An antiderivative of x 2
5 (a)
0 [0, 3] by [ 1, 8] An antiderivative of x 2
3 (a)
0 (x 2 6x 8) dx 6x
F(3) 13
x
3 8 is F(x)
F(0) 6 0 3x 2
6 8x. (x 2 4x) dx 4x is F(x) F(5) F(0) 13
x
3
25
3 2x 2.
0 25
3 Section 5.3
4 (b) Area 5 (x 2 4x) dx 0 30. The region between the graph and the xaxis is a rectangle 4x) dx 4 [F(4) F(0)] [F(5)
25
3 32
3 with a half circle of radius 1 cut out. The area of the region F(4)] 0 32
3 25. An antiderivative of x 2 x. (x 2
3 F( 3) 14
2
2 f (t) dt
1 0 2 1
2 av( f ) F(0) 0) . 2 4
4 . 31. There are equal areas above and below the xaxis. 1) dx 0 1 4 1 1
2 av( f ) 13
x
3 1 is F(x) 1
(1)2
2 is 2(1) 13 3 1 av (x 2 217 1
2 f (t) dt
0 0 0 3
1 (0 32. Since tan is an odd function, there are equal areas above 3 Find x c in [0, 3] such that c2
c2 1
c
1
Since 1 is in [0, 3], x 1. 1 and below the xaxis. 0 33. min f
2 3 x
x
26. An antiderivative of
is F(x)
.
2
6
3
1
x2
1
1
av
dx
[F(3) F(0)]
30
2
3
3
c2
3
Find x c in [0, 3] such that
.
2
2 c2 9
2 1
2 3
2 1
2
8
17
1
2
1
4
8
17
49
68 c 3 Since 3 is in [0, 3], x 27. An antiderivative of
1
1 av 3. 3x 2 1 is F(x) x 3 x. 1 ( 3x 2 1) dx F(1) F(0) 2 0 Find x 3c2 c in [0, 1] such that 3c2 1 2 1 c 1 0 0 1 16
17
0.5 16
17
0.5 1
x4 1 0 1
1 1
(1)
2 dx
1
2 dx x4
1
1
1
2
x4
0.5 1
1
1
dx
x4
0.5 1
1
1
1
4
x4
01
1
1
dx
x4
01
0 35. max sin (x 2 ) 1
3 2 2 f( ) d
/4 1
and max f
2
1
1
dx 1
x4
01 34. f (0.5) 3 /4 1
/2 av( f ) 1 16
2 17 dx
8
17 dx 1
2 8
17 33
34 sin (1) on [0, 1] sin (x 2 ) dx sin (1) 1 0 1 c 36. max 3
1 Since 1 is in [0, 1]. x
3 1
3 av 3 1)2 dx (x . 22 0 3 and min
x 1
(x 1)3.
3
18
1
F(0)]
33
3 37. (b
1 a) min f (x) Find x c in [0, 3] such that (c 1)
1.
c1
1
c 2 or c 0.
Since both are in [0, 3], x 0 or x 2.
29. The region between the graph and the xaxis is a triangle of
height 3 and base 6, so the area of the region
1
2 av( f ) 1
6 9.
2 f (x) dx
4 9
6 3
2 8 8 dx 3 0 on [a, b]
b 0 (b a) min f (x) f (x) dx
a 2 is (3)(6) x 0 1)2 is F(x)
1
[F(3)
3 8
1 3 28. An antiderivative of (x x 38. (b a) max f (x) 0 on [a, b] b f (x) dx
a (b a) max f (x) 0 2 2 on [0, 1] 218 Section 5.4
b 39. Yes, a 46. An antiderivative of F (x) is F(x) and an antiderivative of b av ( f ) dx f (x) dx. a G (x) is G(x). This is because av( f ) is a constant, so
b av( f ) dx a b a b a
b F (x) dx av( f ) b av( f ) a F(b) F(a) G (x) dx av ( f ) x G(b) G(a) a
b (b a)av( f ) (b Since F (x) b a a b F (x) dx G (x) dx, so a b 1 a) G (x), f (x) dx F(b) F(a) G(b) a G(a). b
a f (x) dx s Section 5.4 Fundamental Theorem of
Calculus (pp. 277–288) 40. (a) 300 mi
(b) 150 mi
30 mph 150 mi
50 mph 300 mi
(c)
8h 8h Exploration 1 37.5 mph 2. The function y (d) The average speed is the total distance divided by the
total time. Algebraically,
1 d1
2 t1 d2 d1 d2 t1 t2 1000 m3
100 min
10 m3/min
1000 m3
Time for second release
50 min
20 m3/min
total released
2000 m3
1
Average rate
13 m3/min
total time
150 min
3 41. Time for first release 1 42. 1 sin x dx x dx 0 0 1 43. 1 sec x dx 1 0 0 1 21
x
20 multiples of . There are six of these between 3. In attempting to find F( 10) 10 and 10. tan (t) dt 5, the 3 calculator must find a limit of Riemann sums for the
integral, using values of tan t for t between 10 and 3. The
large positive and negative values of tan t found near the
asymptotes cause the sums to fluctuate erratically so that no
limit is approached. (We will see in Section 8.3 that the
“areas” near the asymptotes are infinite, although NINT is
not designed to determine this.)
4. y tan x 1
2 x2
dx
2 x3 1
60 x 7
6
[1.6, 4.7] by [ 2, 2] 1
bh
2 44. (a) Area 2 tan x has vertical asymptotes at all odd 10 . The driver computed . The two expressions are not equal. t2 Graphing NINT f 5. The domain of this continuous function is the open interval
(b) h2
x
2b C
2 b (c) y (x) dx
0 45. av(x k)
Graph y1 1
k h 2b
x
2b 0 k x k dx 0 hb 2
2b 1
bh
2 11
k
xk 1
kk 1
0 xx 1
and y2
x(x 1) 3
.
2 6. The domain of F is the same as the domain of the
kk 1
k(k 1) x on a graphing calculator and find the point of intersection for x , 1. continuous function in step 4, namely 2 , 3
.
2 7. We need to choose a closed window narrower than
2 , 3
2 to avoid the asymptotes. [1.6, 4.7] by [0, 16]
[1, 3] by [0, 3] 8. The graph of F looks like the graph in step 7. It would be
Thus, k 2.39838 decreasing on 2 , and increasing on vertical asymptotes at x 2 and x 3
.
2 , 3
, with
2 219 Section 5.4
Exploration 2 The Effect of Changing a in 10. x f (t) dt
a dy
dx 1
dx/dy 1
3x Section 5.4 Exercises 1. 3 1. 3 1
dx
x 2
1/2 2x ln x
1/2 (6 [ 4.7, 4.7] by [ 3.1, 3.1] 2. ln 3) 1 ln 1
2 1
2 5 ln 3 ln 5 ln 3 ln 2 5 ln 6 3.208 [ 4.7, 4.7] by [ 3.1, 3.1] 3. Since NINT(x 2, x, 0, 0) 0, the xintercept is 0. 2 4. Since NINT(x , x, 5, 5) 1 0, the xintercept is 5. 5. Changing a has no effect on the graph of y
It will always be the same as the graph of y d
dx 1 1
3x
ln 3 3x dx 2.
2
x 2 11
ln 3 3 f (t) dt.
a 9 26
3 ln 3 f (x). 7.889 x 6. Changing a shifts the graph of y f (t) dt vertically in
a 1 3. a dy
dx dy
2.
dx 3. dy
dx f (t) dt. a 4. 2 5/2
x
5 x 3/2 dx 0 2
32 Quick Review 5.4
1. 5
1 13
x
3 x) dx 0 such a way that a is always the xintercept. If we change
from a1 to a2, the distance of the vertical shift is (x 2 6/5 x 5. dx 5 2
(25
5 0 1/5 5x 1 cos(x 2) 2x 1 6.
2 5 2 x 2 32 5 dx 1 dy
dx 3
3x 5. dy
dx 1 1/2
x
2 5
2
1
2 21 1 0 7. sin x dx cos x 1 0 ( 1) 2 0 8. (1 cos x) dx x sin x
0 2x ln 2 dy
6.
dx 1 1 22.361 2 0 4. 1 5 0) 2 2(tan x)(sec2 x) 2 tan x sec2 x (0 10 1
2 x 2 2
3 0) 2 sin x cos x 2(sec x)(sec x tan x)
2 sec2 x tan x 2
dx
x2 1 1
3 0 1 2x cos (x 2) 2(sin x)(cos x) 1 2 3/2
x
3 7. dy
dx 8. dy
dt
dy
dx 7
7x ( 0 (0 0) 3.142
/3 /3 dx
dt dy/dt
dx/dt 0 2( sin t x sin x cos x
x2 2
5 /6 10. 2y)
y
x (y
1
2y y
2y csc2 3
3 1
x 3.464
5 /6 d cot
/6 3 1) 0) /6 cot t dy
2y
dx 1 2 tan 0 cos t
sin t (1)y d x 2 ( sin x)(x) (cos x)(1)
x2 cos t, 2 sec2 9. 1 9. Implicitly differentiate:
dy
x
dx
dy
(x
dx
dy
dx 0) 2 (
3 3)
3.464
3 /4 3 /4 11. csc x cot x dx csc x ( /4 4 sec x tan x dx
0 ( 2) /3 /3 12. 2) /4 4 sec x 4(2
0 1) 4 0 220 Section 5.4 1 13. (r 1)2 dr 1
(r
3 u 4 1
4 14. 1 0 du 1)3 1 8
3 1 1/2 (u x3 17. Graph y 8
3 0 3x 2 2x. 1) du 0 u 4 2u1/2 u [0, 2] by [ 1, 1]
0 (4
15. Graph y 2 4) (0 0) Over [0, 1]: 0 1 x. (x 3 3x 2 14
x
4 0 x3 14
x
4
1
2 2x) dx x3 1 1
4 0 0 1
4 0 ( 4) 4 4 0 x2 0 1
4 Over [1, 2]:
2 (x 3 3x 2 2x) dx 1 2 Over [0, 2]: (2 2x
2x x) dx 12
x
2
12
x
2 0
3 Over [2, 3]: (2 x) dx 2 Total area
16. Graph y 1
2 2
3x 2 1
4 Total area [0, 3] by [ 2, 3] x3 18. Graph y 2 1
4 x2 2
1 1
4 4x. 2
0
3 3
2 2 2 1
2 5
2 [ 2, 2] by [ 4, 4] 3. Over [ 2, 0]:
0 (x 3 14
x
4 4x) dx 2 2x 2 0
2 Over [0, 2]:
2 [ 2, 2] by [ 4, 10] (x 3 14
x
4 4x) dx 0 Over [ 2,
1 (3x 2 1]:
3) dx Total area
x3 1 3x 2 2 ( 2) 4 2 (b)
(3x 2 3) dx x3 3x 1 1 2 2 (3x 2 Over [1, 2]: 3) dx 1
1 x x3 8 1
is discontinuous at x
1 1, x 1
2 2 ( 2) 4 9
. The area
2 between the graph of f and the xaxis over (1, 3] where 1 1
2 f is positive is (2)(2)
4 1. 1 2 3x 4 0 [ 2, 1) where f is negative is (3)(3) 1 Total area 2 The area between the graph of f and the xaxis over 4 1
2 x2
x 4
x2
x 19. (a) No, f (x) Over [ 1, 1]:
1 4 2x2 4 4 12 3
2 2 x
x 20. (a) No, f (x)
(b) 9 x2
3x 9 9
2 1
dx
1 2.
2 5
2 9 x2
is discontinuous at x
3x 9
3 x
3 ,x 3. 3 Note that f is negative for x in [0, 5]. f (0) 1 and 8
. The area between the graph of f and the
3
1
8
55
xaxis over [0, 5] is (5 0) 1
2
3
6
5
55
9 x2
dx
6
9
0 3x f (5) ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.
 Spring '08
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