3.1 Topics: Antiderivatives and Indefinite Integration Solutions
Find the indefinite integral ( don’t forget the + C ). You may need to rewrite the integrand prior to integrating. Simplify your answer
if possible (do not factor your answer but do rewrite your answer so that there are no negative exponents).
1
1.
  x  3 dx  2 x
2.
x
3.

4.
 2x
4
2
 3x  C
 3x 2  x  4  dx 
1
3
x dx   x 3 dx 
1
3
1 5
1
x  x3  x 2  4 x  C
5
2
3 34
x C
4
1
1 1
1
dx   x 3 dx   x 2  C   2  C
2
2 2
4x
1
1
1
 
 12
6 
2 32
2 32
 x
2
2
2
dx   

 dx    x  6 x  dx  x  6  2 x  C  x  12 x  C
3
3
x
x
 x


x6
5.

6.
  x  1 2 x  3 dx    2 x
7.
 y
8.
 dx   1dx  x  C
9.
  2 cos x  3sin x  dx  2sin x  3cos x  C
2

1
2
1
1
2
5
 5 x  3dx  2  x 3  5  x 2  3x  C  x 3  x 2  3 x  C
3
2
3
2
5
y dy   y 2 y 2 dy   y 2 dy 
2 72
y C
7
10.
  2  csc x cot x  dx  2 x  csc x  C
11.
  tan
12.
  x  x  dx  2 x

2
  1 d    sec2   d  tan   C
4
1
2
 4ln x  C 
1 2
x  ln x 4  C
2
13. A ball is thrown vertically upward from a height of 6 feet with an initial velocity of 60 feet per second. How high will the ball go.
Use a  t   32 ft 2
s
a  t   32 ft 2  v  t    a  t  dt  v  t    32dt  v  t   32t  C
s
since we are told the intital velocity is 6o feet per second we know that v  0   60
v  0   32  0   C  60  C  60 So our velocity function is v  t   32t  60.
At the maximum height the velocity will be zero.... 0  32t  60  t  1.875sec
So if we want to know what the maximum height is we need to find the position
function for this problem and evaluate it at t  1.875sec.
s  t    v  t  dt    32t  60 dt  16t 2  60t  C. So were told that the ball was thrown
from an initial height of 6 feet so we know that when t=0 the position was 6 ft.
6  16  0   60  0   C  C  6. So our position function is s  t   16t 2  60t  6
2
s 1.875   16 1.875   60 1.875   6  62.25 ft
2
14. With what initial velocity must an object be thrown upward (from ground level) to reach the top of the Washington Monument
(approximately 500 feet)?
If the object is thrown upward from ground level we know that its initial position was zero. So our position equation will be
s t   16t 2  v0t  0 . If 500 feet is to be the peak of the object’s path then the velocity at the peak will be zero so
v  t   32t  v0
 0  32t  v0
 t
v0
32
Now we will set s(t)=500 … 500  16t 2  v0 t and set t 
v0
and we will be able to solve for the initial velocity.
32
2
v2 v2
v 
v 
500  16  0   v0  0   500   0  0
64 32
 32 
 32 
v0   32000  178.89
ft
s
 500 
Answer: v0  178.89
v0 2
64
 32000  v0 2
ft
s
15. A baseball is thrown upward from a height of 2 meters with an initial velocity of 10 m . Determine its maximum height.
s
Use a  t   9.8 m 2
s
This problem is similar to 13 except the units are different.
a  t   9.8 m 2  v  t    a  t  dt  v  t    9.8dt  v  t   9.8t  C
s
m
Since the initial velocity is 10 we have v  0   9.8  0   C  10  C  10
s
v  t   9.8t  10  s  t    v  t  dt  s  t     9.8t  10 dt
s  t   4.9t 2  10t  C Since the initial height is 2 meters we have 2  4.9  0   10  0   C  C  2
2
so s  t   4.9t 2  10t  2. At the maximum height the velocity will be zero... v  t   9.8t  10  0
which gives us a time of t 
2
10
9.8
 10 
 10 
 10 
s
  4.9 
  10 
  2  7.1m
 9.8 
 9.8 
 9.8 