Math 1152
Assignment 1 (Solutions)
Math 1152
Assignment 1 (Solutions)
(Due date: Jan. 20, 2017)
Page 272:
6. The general antiderivative of
fx  2x 3  x 2  5x
is
Fx  1 x 4  1 x 3  5 x 2  C.
2
3
2
10. The general antiderivative of
fx  x 2  22  33
x
x
is
Fx  2x  3 2  1 x 3  C.
3
2x
18. The general antiderivative of
fx 
1
1  3x
is
Fx  1 ln x  1  C,
3
3
or
Fx  1 ln|3x  1|  C.
3
20. The general antiderivative of
fx  e x/2  e x/2
is
1
1
Fx  2e 2 x  2e  2 x  C.
26. The general antiderivative of
fx  cos3x
is
Fx  1 sin3x  C.
3
30. The general antiderivative of
fx  3 sin  x  4 cos   x
3
4
is
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Math 1152
Assignment 1 (Solutions)
9 cos  x  16 sin   x  C.
Fx  

3
4
42. The general antiderivative of
fx  sin 2 a 2 x  1
is
Fx  1 x  1 2 sin 2 2a 2 x  2  C.
2
4a
50. The general solution for differential equation
dy
 e 4x , x  0
dx
is
y   1 e 4x  C.
4
60. First we find the general solution for differential equation
2
dy
 x , x0
3
dx
is
y  1 x 3  C.
9
Then we use the initial condition:
y  2 when x  0
and have C  2. Therefore the solution is
y  1 x 3  2.
9
70. First we find the general solution for differential equation
dT  cost, t  0
dt
is
1 sint  C.
T 
Then we use the initial condition:
T0  3
and have C  3. Therefore the solution is
1 sint  3.
Tt  
Page 291
8. Write the sum in expanded form
3

k1
k2  12  22  32  1  4  9 .
5
2
10
k 1
12  1
22  1
32  1
2
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Math 1152
Assignment 1 (Solutions)
10. Write the sum in expanded form
4
 kx  0x  1x  2x  3x  4x
k0
 1x  2x  3x  4x.
16. Write sum in sigma notation
4

1  1  1  1 
1
2
3
4
1 .
k
k1
20. Write sum in sigma notation
n

1  1  1  1 
1
2
4
2n
k0
1 .
2k
n
28. Use the algebraic rules and the formula for  k1 k 2 , we have
n
n
k1
k1
n
k  2k  2  k 2  2 2 

n
 k   22
2
k1
k1
nn  12n  1

 4n
6
30. Expand the summation notation, we have
10
1 k  1 0  1 1  1 2  1 3  1 4  1 5
k0
 1 6  1 7  1 8  1 9  1 10
 1  1  1  1  1  1  1  1  1  1  1  1.
34. For n  5, we have the partition P on interval 1, 1 :
1  1
x  x i 
 0. 4
5
x 0  1,
x 1  1  0. 4  0. 6,
x 2  0. 6  0. 4  0. 2,
x 3  0. 2  0. 4  0. 2,
x 4  0. 2  0. 4  0. 6,
x 5  1.
Use right endpoint we have
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Math 1152
Assignment 1 (Solutions)
c 1  0. 6  1, 0. 6,
c 2  0. 2  0. 6, 0. 2,
c 3  0. 2  0. 2, 0. 2,
c 4  0. 6  0. 2, 0. 6,
c 5  1  0. 6, 1.
We have the Riemann sum
5
5
k1
k1
 fc k x k  0. 4 2  c 2k 
 0. 42  0. 6 2   2  0. 2 2   2  0. 2 2   2  0. 6 2   2  1 2 
 0. 4  11. 8  4. 72.
Therefore we have
1
 1 2  x 2 dx  4. 72
44. If P is a partition of 0, , then
n

P0
lim
k1
1 x 
ck  1 k

0
1 dx.
x1
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