1.
If an object in motion on the x axis has velocity v(t) and position F (t), then the displacement of the object
between a ≤ t ≤ b is the NET change from time a to time b. In other words,
Z b
Displacement from a to b =
v(t) dt = F (b) − F (a).
a
Notice that the position of the object in motion is given by F (t) =
R
v(t) dt.
Assume that the object in motion has velocity v(t) = 5t4 + 3t2 whose units are meters per second.
a)
How far does the object travel between t = 1 and t = 3?
Solution: From the formula above we have that the displacement is
Z 3
3
5t4 + 3t2 dt = t5 + t3 1 = (35 + 33 ) − (15 + 13 ) = 35 + 33 − 2
meters.
1
b)
What is the position of the object after 3 seconds?
Solution: The position of the object after 3 seconds is
F (3) = 35 + 33
2.
meters away from the starting point.
Find
Z
a)
cos(ln(x))
dx.
x
Solution: Technique: Integration by Substitution. Take u = ln(x), so du =
Z
Z
cos(ln(x))
dx = cos(u) du = sin(u) + C = sin(ln(x)) + C.
x
Z
b)
tan4 (x) sec2 (x) dx.
c)
dx
x .
Thus
Solution: Technique: Integration by Substitution. Take u = tan(x), so du = sec2 (x) dx. Thus
Z
Z
tan5 (x)
u5
4
2
+C =
+ C.
tan (x) sec (x) dx = u4 du =
5
5
Z
x−1
dx.
1 + x2
Solution: First notice that
Z
Z
Z
x−1
x
1
dx
=
dx
−
dx.
1 + x2
1 + x2
1 + x2
The first integral on the right can be solve by substitution. Take u = 1 + x2 , so du = 2x dx. Thus
Z
Z
x
1
1
1
dx =
u−1 du = ln |u| + C = ln(1 + x2 ) + C.
2
1+x
2
2
2
The second integral on the right is in our basic list
Z
1
dx = arctan(x) + C.
1 + x2
Hence
Z
x−1
dx =
1 + x2
Z
x
dx −
1 + x2
Z
1
1
dx = − ln(1 + x2 ) − arctan(x) + C.
2
1+x
2
Z
d)
ln(2x + 1) dx.
Solution: Technique: Integration by Parts. Take u = ln(2x + 1) nd dv = dx, so du =
v = x. Thus
Z
Z
Z
2x
2x + 1 − 1
ln(2x + 1) dx = x ln(2x + 1) −
dx = x ln(2x + 1) −
dx
2x + 1
2x + 1
Z
Z
Z 1
2
1
dx = x ln(2x + 1) − dx +
dx
= x ln(2x + 1) −
1−
2x + 1
2
2x + 1
ln(2x + 1)
(2x + 1)
= x ln(2x + 1) − x +
+C =
ln(2x + 1) − x + C.
2
2
3.
2
2x+1
dx and
Evaluate
Z
2
√
a)
0
x
dx
1 + 2x2
Solution: We find first the indefinite integral by of integration by substitution. Take u = 1 + 2x2 so
dx = 4x dx. Then
9
Z 2
Z
√
x
1 √
1 u(2) −1/2
1 1/2 √
= ( 9 − 1) = 1.
dx =
u
u
du =
2
4
2
2
1 + 2x
0
u(0)
1
Z
3
b)
1
e3/x
dx
x2
Solution: We find first the indefinite integral by of integration by substitution. Take u =
du = − x32 dx. Then
1
Z
Z 3 3/x
1
1 u(3) u
1 u e
3
dx
=
−
e
du
=
−
(e
)
= − 3 (e − e ).
2
x
3
3
u(1)
1
3
Z
c)
3
x,
so
π
t sin(3t) dt.
0
Solution: We find first the indefinite integral by means of integration by parts. Take u = t and
. Then
dv = sin(3t) dt, so du = dt and v = − cos(3t)
3
π
π
Z π
Z π
cos(3t) 1
cos(3t) 1
t sin(3t) dt = −t
+
cos(3t)
dt
=
−t
+
sin(3t)
3
3
3
9
0
0
0
0
=
−π
cos(3π) 1
cos(0) 1
π
+ sin(3π) − −0
+ sin(0) = .
3
9
3
9
3