This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 5]. (b) What does the Mean Value Theorem for integrals say? Find a c in [2, 5] such that fave = f (c)
where f (x) = (x 3)2
Solution: (a) Note that
Z5p
Z
1
1 1 26 p
1
fave =
t 1 + t2 dt {z}
=
·
u du =
5 00
521
10
2 2 3/2
u
3 u=1+t 26
1 ! = 1
(263/2
15 1) (b) Mean Value Theorem for Integrals: If f is continuous on [a, b], then there exists a number c in
[a, b] such that
Zb
1
f (c) = fave =
f (x)dx.
b aa
Now note that
Z5
Z
1
1 22
1 u3 2
9
fave =
(x 3)2 dx {z}
=
u du = ·
= =1
5 22
31
3319
u= x 3 3)2 = 1 and so c = 3 ± 1 = 2, 4.
Z cos(y) p
Zx
cos(t)
3
2 dt (b) y =
4. Find the derivative of : (a) g (y ) =
1t
dt
p
t
1
x
Now f (c) = fave implies (c Solution: (a) g 0 (y ) =
(b) p
3
1 cos2 (y ) · dy
d
=
dx
dx Z x p
sin(y ) 3 1 sin(y ) = cos(t)
dt
t Z 0 cos(t)
dt +
t Z x cos(t)
dt
t
x
x
0
" Zp
#
Zx
x
d
cos(t)
cos(t)
=
dt +
dt
dx
t
t
0
0
p
p
cos( x) 1
cos(x)
cos( x) cos(x)
p
p+
=
=
+
x
2x
x
x 2x
Zx
1
5. Find the intervals where the function f (x) =
dt is concave up and concave down.
2...
View
Full
Document
This note was uploaded on 02/04/2014 for the course APPM 1350 taught by Professor Segur,harv during the Fall '07 term at Colorado.
 Fall '07
 SEGUR,HARV

Click to edit the document details