MATH 101 V2A
January 14th – Practice problems
Solutions
1. Differentiate the following functions:
Z x
2
e−t dt.
(a)
1
x
Z
2
2
e−t dt and let f (t) = e−t . Then the FTC tells us that
Solution: Let F (x) =
1
2
F 0 (x) = f (x) = e−x .
Z
1
(b) sin(x)
2
e−t dt.
x2
b
Z
Z
f (t)dt = −
Solution: Using the property of integrals that
a
Z
1
−t2
e
x2
Z
x2
a
f (t)dt, we see that
b
2
e−t dt = −F (x2 ),
dt = −
1
where F (x) is defined as in part (a). So, using the Product and Chain Rules, we
get
Z 1
d
d
−t2
e dt =
sin(x)
− sin(x)F (x2 )
dx
dx
x2
d
d
2
2
=−
sin(x) F (x ) − sin(x)
F (x )
dx
dx
= − cos(x)F (x2 ) − sin(x)F 0 (x2 ) · 2x.
2
4
From part (a) we know that F 0 (x) = e−x , so F 0 (x2 ) = e−x . Therefore
d
dx
Z
1
sin(x)
e
x2
−t2
Z
dt = − cos(x)
1
x2
2
4
e−t dt − sin(x)e−x · 2x.
Z
x3
(c) sin(x)
2
e−t dt.
x2
Solution: For this problem we need to use another property of integrals: If f (t) is
integrable on an interval I containing the points a, b, c, then
Z b
Z c
Z b
f (t)dt =
f (t)dt =
f (t)dt
a
a
c
(Note: c does not have to be between a and b for this property to hold).
2
Now, the function f (t) = e−t is integrable on every interval (because it is continuous everywhere), so I can choose a = x2 , b = x3 and c = 1 and use the above
property to get
Z x3
Z 1
Z x3
2
−t2
−t2
e dt =
e dt +
e−t dt,
x2
x2
1
and therefore
Z
x3
sin(x)
−t2
e
Z
1
dt = sin(x)
x2
e
−t2
Z
dt + sin(x)
x2
x3
2
e−t dt,
1
We know what the derivative of the first term is from part (b), so we just need to
calculate the derivative of the second term.
Z x3
2
e−t dt = F (x3 ), where F (x) is again defined as in part (a).
Now, we can write
1
Then
d
dx
Z
sin(x)
x3
!
−t2
e
1
dt
=
d
d
3
3
sin(x) F (x ) + sin(x)
F (x )
dx
dx
= cos(x)F (x3 ) + sin(x)F 0 (x3 ) · 3x2 .
2
6
Since F 0 (x) = e−x (from part (a)), F 0 (x3 ) = e−x , and we see that
!
Z x3
Z x3
d
2
2
6
−t
e dt = cos(x)
et dt + sin(x)e−x · 3x2 .
sin(x)
dx
1
1
Therefore, (using part (b))
!
Z x3
Z x2
d
2
2
4
−t
sin(x)
e dt = − cos(x)
e−t dt − sin(x)e−x · 2x
dx
x2
1
Z x3
2
6
+ cos(x)
et dt + sin(x)e−x · 3x2 .
1
2
2. Evaluate the following integrals.
Z b
t2 dt.
(a)
0
Solution: One antiderivative of t2 is F (t) = 13 t3 (check: F 0 (t) = t2 ). So, the FTC
tells us that
Z b
1
t2 dt = F (b) − F (0) = b3 .
3
0
Z
1
(t3 + 3t − 2)dt.
(b)
−2
Solution: One antiderivative of t3 + 3t − 2 is the function F (t) = 41 t4 + 32 t2 − 2t, so
the FTC tells us that
Z 1
1 3
1
3
57
3
4
2
(t +3t−2)dt = F (1)−F (−2) =
+ − 2 − (−2) + (−2) − 2(−2) = − .
4 2
4
2
4
−2
Z
(c)
π
sin(t)dt.
0
Solution: One antiderivative of sin(t) is F (t) = − cos(t), so the FTC tells us that
Z π
sin(t)dt = F (π) − F (0) = (− cos(π)) − (− cos(0)) = 2.
0
3