7.1, #21.
|
|
ó
õ
|
|
æ
ç
è
|
|
3
t
|
- |
2
t2
|
|
ö
÷
ø
|
dt = 3ln|t|+2t-1+C |
|
7.1, #26.
7.1, #43.
The indefinite integral is 7.1, #14 on assignment 6.
|
|
ó
õ
|
2
1
|
|
1+y2
y
|
dy = (lny+y2/2) |
ê
ê
|
2
1
|
= ln2+3/2 = 2.193 |
|
7.1, #54.
The average value of sint on the interval 0 £ t £ 2p is
|
|
1
2p-0
|
|
ó
õ
|
2p
0
|
sint dt = 0. |
|
We can evaluate this integral using an antiderivative or by knowing
that the areas above and below the x-axis are equal.
This is becasue sin(t+p) = -sint.
The part of the graph above the x-axis balances the part below.
This is a huristic argument that the average should be 0.
The average value of sint on 0 £ t £ p is
|
|
1
p
|
|
ó
õ
|
p
0
|
sint dt = |
1
p
|
[-cost]0p = |
1
p
|
(-(-1)-(-1)) = |
2
p
|
. |
|
7.2, #2. Use the substitution u = x2, du = 2x dx to get
|
|
ó
õ
|
2xcos(x2) dx = |
ó
õ
|
cosu du = sinu+C = sin(x2)+C |
|
7.2, #5.
Use the substitution u = sinx, du = cosx dx to get
|
|
ó
õ
|
esinx cosx dx = |
ó
õ
|
eu du = eu+C = esinx+C |
|
7.2, #6.
Use the substitution u = x2+1, du = 2x dx to get
|
|
ó
õ
|
|
x
x2+1
|
dx = |
1
2
|
|
ó
õ
|
|
du
u
|
= |
1
2
|
ln|u|+C = |
1
2
|
ln(x2+1)+C |
|
File translated from TEX by TTH, version 2.00.
On 14 Aug 1999, 21:19.