Part II: The Answers
360
Answers
501–600
Then flip the limits of integration and change the sign of the first integral:
f x
t
t
dt
t
t
dt
t
t
dt
t
t
x
x
x
( ) =
−
+
+
−
+
= −
−
+
+
−
+
∫
∫
∫
2
4
2
0
2
4
0
6
2
4
0
2
2
4
1
1
1
1
1
1
1
1 1
0
6
dt
x
∫
Note also that to find d
dx
g t dt
a
h x
( )
( )
∫
, you can use the substitution u = h(x) and then
apply the chain rule as follows:
d
dx
g t dt
d
dx
g t dt
d
du
g t dt du
dx
g u du
d
a
h x
a
u
a
u
( )
( )
( )
( )
( )
∫
∫
∫
(
)
= ′
(
)
=
=
x x
g h x
du
dx
= ′ (
)
(
)
( )
All this tells you to substitute the upper limit of integration into the integrand and multiply by the derivative of the upper limit of integration. Therefore, the derivative of
f x
t
t
dt
t
t
dt
x
x
( ) = −
−
+
+
−
+
∫
∫
2
4
0
2
2
4
0
6
1
1
1
1
is
′
= −
−
+
+
−
+
= −
f x
x
x
x
x
( )
( )
( )
( )
( )
( )
( )
2
1
2
1
2
6
1
6
1
6
2 4
2
4
2
4
x x
x
x
x
2
4
2
4
1
16
1
6 36
1
1 296
1
−
+
+
−
+
,
557.
2 5
1
5
2
x
x
( ) − ln
Part of the fundamental theorem of calculus states that if the function g is continuous
on [a, b], then the function f defined by
f x
g t dt
a x b
a
x
( )
( )
=
≤ ≤
(
)
∫
where
is continuous on [a, b] and is differentiable on (a, b). Furthermore, f '(x) = g(x).
To use the fundamental theorem of calculus, you need to have the variable in the
upper limit of integration. Therefore, to find the derivative of the function
f x
dt
t
x
x
( ) log
= ∫ 5
5
2
, first split the integral into two separate integrals:
f x
dt
dt
dt
t
x
x
t
x
t
x
( ) log
log
=
=
+
∫
∫
∫
5
5
5
5
2
5
2
0
0
Then flip the limits of integration and change the sign of the first integral:
f x
dt
dt
dt
dt
t
x
t
x
t
x
t
x
( ) log
log
=
+
= −
+
∫
∫
∫
∫
5
5
5
5
5
2
5
2
0
0
0
0
360
Answers
501–600
Then flip the limits of integration and change the sign of the first integral:
f x
t
t
dt
t
t
dt
t
t
dt
t
t
x
x
x
( ) =
−
+
+
−
+
= −
−
+
+
−
+
∫
∫
∫
2
4
2
0
2
4
0
6
2
4
0
2
2
4
1
1
1
1
1
1
1
1 1
0
6
dt
x
∫
Note also that to find d
dx
g t dt
a
h x
( )
( )
∫
, you can use the substitution u = h(x) and then
apply the chain rule as follows:
d
dx
g t dt
d
dx
g t dt
d
du
g t dt du
dx
g u du
d
a
h x
a
u
a
u
( )
( )
( )
( )
( )
∫
∫
∫
(
)
= ′
(
)
=
=
x x
g h x
du
dx
= ′ (
)
(
)
( )
All this tells you to substitute the upper limit of integration into the integrand and multiply by the derivative of the upper limit of integration. Therefore, the derivative of
f x
t
t
dt
t
t
dt
x
x
( ) = −
−
+
+
−
+
∫
∫
2
4
0
2
2
4
0
6
1
1
1
1
is
′
= −
−
+
+
−
+
= −
f x
x
x
x
x
( )
( )
( )
( )
( )
( )
( )
2
1
2
1
2
6
1
6
1
6
2 4
2
4
2
4
x x
x
x
x
2
4
2
4
1
16
1
6 36
1
1 296
1
−
+
+
−
+
,
557.
2 5
1
5
2
x
x
( ) − ln
Part of the fundamental theorem of calculus states that if the function g is continuous
on [a, b], then the function f defined by
f x
g t dt
a x b
a
x
( )
( )
=
≤ ≤
(
)
∫
where
is continuous on [a, b] and is differentiable on (a, b). Furthermore, f '(x) = g(x).
To use the fundamental theorem of calculus, you need to have the variable in the
upper limit of integration. Therefore, to find the derivative of the function
f x
dt
t
x
x
( ) log
= ∫ 5
5
2
, first split the integral into two separate integrals:
f x
dt
dt
dt
t
x
x
t
x
t
x
( ) log
log
=
=
+
∫
∫
∫
5
5
5
5
2
5
2
0
0
Then flip the limits of integration and change the sign of the first integral:
f x
dt
dt
dt
dt
t
x
t
x
t
x
t
x
( ) log
log
=
+
= −
+
∫
∫
∫
∫
5
5
5
5
5
2
5
2
0
0
0
0