Download Solution

Math 3: Fundamental theorem of calculus
1. Compute the following derivatives.
Z x
d
2
(a)
t tan(t)dt
dx
0
Solution: By the fundamental theorem,
Z x
d
2
t tan(t)dt = x2 tan(x).
dx
0
d
(b)
dx
3
Z
ln(t) dt
2
x
Solution: In order to use the fundamental theorem, we first have to switch the endpoints of integration, getting that
Z 3
Z x
d
d
ln(t)2 dt = −
ln(t)2 dt = − ln(x)2 .
dx
dx
x
3
d
(c)
dx
Z
!
x2
sin(t)et dt
0
Solution: If g(x) =
rule, we get that
Rx
0
sin(t)et dt, then we are looking for
d
dx
d
(d)
dx
Z
Z
d
2
dx (g(x )).
Using the chain
!
x2
t
sin(t)e dt
2
= 2x sin(x2 )ex .
0
!
x3
2
cos (t)dt
sin(x)
Solution: We first split the integral as
!
!
Z x3
Z 0
d
d
d
2
2
cos (t)dt =
cos (t)dt +
dx
dx
dx
sin(x)
sin(x)
Z
!
x3
2
cos (t)dt .
0
We could have used any number in the domain of cos2 (x) in place of 0. We now
proceed as in (b) and (c), flipping the first integral and applying the chain rule to
both. We ultimately get that
!
Z x3
d
cos2 (t)dt = − cos(x) cos2 (sin(x)) + 3x2 cos2 (x3 ).
dx
sin(x)
2. Compute the following definite integrals.
5
Z
(3x2 + 4x − 7)dx
(a)
1
Solution: If f (x) = 3x2 + 4x − 7, then F (x) = x3 + 2x2 − 7x is an antiderivative of
f (x). By the fundamental theorem of calculus,
Z
5
(3x2 + 4x − 7)dx = F (5) − F (1) = 144.
1
4
Z
(b)
2
x3 + x + 1
dx
x2
Solution: First we need to rewrite the integrand as a function that we know an
antiderivative for, getting
Z 4 3
Z 4
x +x+1
1
1
dx =
x + + 2 dx.
x2
x x
2
2
We then calculate that
2
4
Z 4
25
1
1
x
1
25
=
x + + 2 dx =
+ ln(x) −
+ ln(4) − ln(2) =
+ ln(2).
x x
2
x 2
4
4
2
3. Find a function f (x) and a number a such that
Z x
f (t)
2+
dt = 5x−1 .
6
t
a
Solution: To find f (x), we differentiate both sides of the above equation, getting that
f (x)
= −5x−2 ,
x6
hence f (x) = −5x4 . We substitute this expression for f (x) back into the original equation,
getting
Z x
2+
−5t−2 dt = 5x−1 .
a
We now use the fundamental theorem of calculus to evaluate the integral on the left. We
Page 2
now have
Z
2+
x
−5t−2 dt = 5x−1
x
2 + 5t−1 a = 5x−1
a
2 + 5x−1 − 5a−1 = 5x−1
5
a= .
2
Page 3