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(b) (c) (d) (e) (f) 3 (a) 3 3 ln x 2 dx = ln2 x 2 + C x 4 1 sin x cos x dx = sin2 x + C 2 1 tan x ln cos x dx = ln2 cos x + C 2 sec2 x etan x dx = etan x + C (2x + 1) (x 2 + x + 7)43 dx = We have
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= lim = lim = lim =2+ = (b) We have x2 dx = lim k 3 1 = lim
j =1
j2 2 3j + 2 + 3 k k k 2k 3 + 3k 2 + k 1 2 k2 + k 3 + 2+ 3 k 2 6 k k 3 1 1 1 3 + + + + 2 2k 3 2k 6k 2
3 1 + 2 3
23 6
j =1 k
1 + 3
2j k
j =1
4j 2 4j 2 + 2 1 3k k k
Solutions to Exercises
= lim
j =1
8j 8j 2 2 + 2 3 3k 3k 3k
2 k 2 + k 8 2k 3 + 3k 2 + k 8 + 2+ 3 k 3k 2 6 3k 3k 2 4 16 = + 3 3 18 2 = 9 3 3 x3 2 3 4 x + 7x x 4x + 7 dx = 3 1 1 = lim k 13 14 + 7 1 3 172 = (9 81 + 21) (1/3 1 + 7) = 3 = 33 34 + 7 3 3
AM FL Y
sin x cos x dx =
sin2 x ex 2 2
e36 sin2 6 2 2
e4 sin2 2 2 2
ln x + x sin x 2 dx = x ln 4 + 2
cos x 2 ln2 x + 2 2
cos 42 2
ln 1 cos 12 + 2 2
ln2 4 cos 16 cos 1 + 2 2 2 sin x 3 3
tan x x 2 cos x 3 dx = ln | cos x| ln | cos 2| sin 23 3
ln | cos 1|
sin 13 3
= ln | cos 2|
ln x 2 dx = x
sin 1 sin 8 + ln | cos 1| + 3 3 e ln2 x 2 ln2 e2 ln2 12 = 1 0 = 1 = 4 4 4
4
x cos x sin x dx =
2 3 3
sin2 x 3 6
= 5 (a)
sin2 83 6
sin2 43 6
sin 512 sin2 64 6 6
Area = =
x3 x2 + + 6x x + x + 6 dx = 3 2
53 52 + +6 5 3 2
/4 0
2 3 22 405 + +6 2 = 3 2 6 sin2 x 2
Area = =
sin x cos x dx =
1 sin /4 sin 0 = 2 2 4
Area =
dx =
ex 2
e1 e4 e e2 = = 2 2 2 2
(d) 6
Area =
ln x dx = x
ln x 2
1 0 1 ln2 e ln2 1 = = 2 2 2 2 2
positive for x (k /6, (k + 1) /6), k even negative for x (k /6, (k + 1) /6), k odd
Fig S46(b) Fig S46(a)
Solutions to Exercises
Fig S46(c) Fig S46(d)
0 2 2 0
Area =
(x 3 + 3x) dx + x 4 3x 2 + 4 2
0 2
x 3 + 3x dx x 4 3x 2 + 4 2
0 0 ( 2)4 3 ( 2)2 + = + 4 2 4 2 + 24 3 2 2 0 0 + 4 2 4 2
11 (j +1) /6
= 20
Area = = =
j = 12
( 1)j sin 3x cos 3x dx ( 1)j 1 sin 6x dx 2
(j +1) /6 j /6
j = 12 j /6 11 (j +1) /6 j = 12 j /6 11
( 1)j 2
cos 6x 6
4
j = 12
1 1 6 6
= 8 (c) Area =
1 1/2 2
ln x dx + x
ln x dx x
ln x = 2
ln2 x + 2
ln (1/2) 0 = 2 2 07404 (d) Area =
0 3 3 0 x4 0 3
12 02 2 2
x 3 ex dx + e + 4 e81
x 3 ex dx
e = 4
1 = 4 4 8 (a) Signed Area = = (b) Signed Area = = (c) Signed Area = = 12
2 2
e81 1 4 4
e81 1 2 2
2 2
x 3 + 3x dx = + 3 22 2
x 4 3x 2 + 4 2 ( 2)4 4 +
2 2
24 4
2 2
3 ( 2)2 2
= 0
sin 3x cos 3x dx =
2 2
1 sin 6x dx 2 = 0
cos 6x 1 2 6
e 1/2
1 1 = 6 6
ln x dx = x
ln2 x 2
ln (1/2) 2 2 02598
Signed Area = Area =
1 1 3 3 3 x4
Solutions to Exercises
(d) 9 (a) x e dx = ex 4
e81 e81 = 0 4 4
1 1
[ 3x 2 + 10] [2x 2 4] dx =
1 1
5x 2 + 14 dx
5x 3 + 14x = 3
5 + 14 3
5 74 14 = 3 3
x3 x4 Area = x x dx = 3 4 0 0 1 1 0 0 1 = = 3 4 3 4 12
Area =
1 3
[ x 2 + 3] 2x dx =
x3 + 3x x 2 3
1 3
1 27 + 3 ( 3) ( 3)2 = + 3 1 12 3 3 32 = 3 e e x2 [x ln x x] x ln x dx = Area = 2 1 1 = e2 [e 1 e] 2
/4 0
12 e2 3 [1 0 1] = 2 2 x2 + cos x 2 2
/4 0
Area = =
x sin x dx =
2 + 32 2
0 2 +1 = + 1 2 32 2
x2 Area = e x dx = e 2 0 0 9 0 11 = e3 e0 = e3 2 2 2 Area =
x2 x3 x x dx = 2 3
1 1 2 3
0 0 2 3
1 = 6
5
(b) Area =
0 1
x x 2 dx = x 3/2 x 3 3/2 3 1 = 3
= (c)
3 3 ( 3)5 x5 2 4 3 Area = 3x x dx = x = ( 3) 5 3 5 3 12 3 ( 3)5 3 = ( 3) 5 5 3
2 1 3 3
0 0 3 3
Area = =
1 1
[ 2x 2 + 3] x 4 dx =
x5 2x 3 + 3x 3 5 = 64 15
1 1
1 2 +3 3 5
21/4 4
( 1) 2 3 3 5
2x 5 Area = [ x + 2] [x 2] dx = 4x 5 21/4 = 4 21/4 = 32 21/4 5
1 3
21/4 21/4
2(21/4 )5 5
4 ( 21/4 )
2( 21/4 )5 5
Area =
[ x 2 + 3] 2x dx =
x3 + 3x x 2 3
1 3
1 27 + 3 ( 3) ( 3)2 = + 3 1 12 3 3 = 32 3
5
1 (a) limx 0 (cos x 1) = 0 and limx 0 x 2 x 3 = 0 so l H pital s Rule applies Thus cos x 1 sin x = lim 2 x3 x 0 x x 0 2x 3x 2 lim
Solutions to Exercises
Now l H pital s Rule applies again to yield = lim (b) 1 cos x = x 0 2 6x 2
limx 0 e2x 1 2x = 0 and limx 0 x 2 + x 4 = 0 so l H pital s Rule applies Thus 2e2x 2 e2x 1 2x = lim x 0 x 0 2x + 4x 3 x2 + x4 l H pital s Rule applies again to yield lim 4e2x = 2 x 0 2 + 12x 2 limx 0 cos x = 0, so l H pital s Rule does not apply In fact the limit does not exist limx 1 [ln x]2 = 0 and limx 1 (x 1) = 0 so l H pital s Rule applies Thus = lim [2 ln x]/x [ln x]2 = 0 = lim x 1 (x 1) x 1 1 lim
(c) (d)
limx 2 (x 2)3 = 0 and limx 2 sin(x 2) (x 2) = 0 so l H pital s Rule applies Thus (x 2)3 3(x 2)2 = lim x 2 cos(x 2) 1 x 2 sin(x 2) (x 2) lim Now l H pital s Rule applies again to yield = lim 6(x 2) x 2 sin(x 2)
We apply l H pital s Rule one last time to obtain = lim (f) 6 = 6 x 2 cos(x 2)
limx 1 (ex 1) = 0 and limx 1 (x 1) = 0 so l H pital s Rule applies Thus ex 1 ex = lim = e lim x 1 x 1 x 1 1 limx + x 3 = limx + (ex x 2 ) = + so l H pital s Rule applies Thus x3 3x 2 = lim x x + ex x 2 x + e 2x lim
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