9·3
5
7. By Property 5,
¯ [ f (x ) g (x )]dx ¯
8. By Property 5,
¯ [ f ( x ) g ( x )]dx ¯
9. By Property 4,
¯
10. By Property 4,
¯
1
5
1
5
1 5
1
5
5
f ( x ) dx ¯ g ( x ) dx 8 22 14.
1 5
1
1
5
f ( x ) dx ¯ g ( x ) dx 8 22 30. 1
1 1 5 1 f ( x ) dx ¯ f ( x ) dx ( 8) 4. 2 2 1 2 5
5
5
2 g ( x ) dx ¯ 3 f ( x ) dx 2 ¯ g ( x ) dx 3 ¯ f ( x ) dx 2(22) 3( 8) 20. 1
1
1
1.
1 d § x 2 (t 3) 5 dt ¶ ( x 2 3) 5 2 ¯ ¨ · 0 © ¸ dx ( x 3)5
2.
d § x 3t 5 dt ¶ 3x 5 ·¸ dx ¨© ¯1
3.
d § x4 ¶ t sin t dt · ( x 4 )sin( x 4 )(4 x 3 ) 4 x 7 sin( x 4 ) dx ¨© ¯P ¸
4.
d § 5x2 3 2 ¶ 3 t dt · (5 x 2 )2 (10 x ) 10 x 3 25 x 4 10 x 2 3 25 x dx ¨© ¯ 5 ¸
5.
d § x 2 2 (t 2t 1) dt ¶ ( x 2)2 2( x 2) 1 x 2 2 x 1 ·¸ dx ¨© ¯ 10 x
6. If F ( x ) ¯ sin(3t ) dt , then F `( x ) sin(3x ). 0
7. If F ( x ) ¯
4x
5
9·4
4 1 1 dt , then F `( x ) 4 4x 1 . t 1 (4 x ) 1
sin x
8. If F ( x ) ¯
0
9. If F ( x ) ¯
3
10. If F ( x ) ¯
8
x
6t 2 dt , then F `( x ) 6(sin x )2 cos x 6 sin 2 x cos x .
3 1 1 2t 4 dt , then F `( x ) 2( x )4 x 2 x 2 . 2
2 x 1
(3t 7 ) dt , then F `( x ) (3(2 x 1) 7 ) 2 12 x 8.
Note: In these exercises, the symbol means “implies.” 1. By the Mean Value Theorem, you have 2
¯
1 1
(2 x 6) dx (2c 6)(1 ( 1)) ( x 2 6 x )
1 1
(2c 6)(2)
2
((1) 6(1)) (( 1) 6( 1)) 4c 12 12 4c 12 0 c . 4
¤ 10 3 ³ ¯0 (2 5 x )dx (2 5 c )(4 0) ¥¦ 2x 3 x 2 ´µ 0 ¤ 10 32 ³ 16 10 32 ³ ¤ 80 (2 5 c )(4 ) ¥ 2(4 ) 4 ´ ¥ 2(0) 0 ´ 8 20 c 8 8 20 c c. 3 9 3 3 µ µ ¦ ¦ 4
2. By the Mean Value Theorem, you have
4
3. By the Mean Value Theorem, you have
¯
4
1
¤ 4³ ¤ 2³ ¤ 4³ 4 dx ¥ 3 ´ (4 1) ¥ 2 ´ ¥ 3 ´ (33) 3 ¦ x µ1 ¦c µ ¦c µ x
¤ 2 ³ ¤ 2 ³ 12 15 12 4 3 ¥¦ 4 2 ´µ ¥¦ 12 ´µ c 3 8 c 3 2 5 c. 4. By the Mean Value Theorem, you have
¯
P 0
P
sin x dx (sin c )(P 0) ( cos x ) 0 (sin c )(P )
( cos P ) ( cos 0) P sin c (sin c )(P ) ( ( 1)) ( (1)) P sin c 2 P sin c ¤ 2³ sin c sin 1 ¥ ´ c . ¦Pµ
140
Worked solutions
2 P