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MAT126 Fall 2009
Practice Midterm II
The actual midterm will consist of six problems
1
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View Full Document Problem 1
If the function
g
(
x
) is given by
g
(
x
) =
x
2
Z
2
x
t
ln
tdt,
compute the derivative
g
0
(
x
)
a) by using the Fundamental Theorem of Calculus to diﬀerentiate the inte
gral
b) by using the Evaluation Theorem to ﬁrst evaluate
g
(
x
) explicitly, and
then diﬀerentiating.
Solution:
a) We note that
g
(
x
) has the form similar to the fundamental theorem of
calculus, but the limits of integration are not as required to apply the theorem.
To apply the fundamental theorem of calculus, we denote
F
(
x
) =
Z
x
0
t
ln
tdt,
so that then
g
(
x
) =
F
(
x
2
)

F
(2
x
) (recall that the integral of a function from
2
x
to
x
2
is equal to the diﬀerence of the integrals from 0 to
x
2
and from 0 to
2
x
).
By the fundamental theorem of calculus we have
F
0
(
x
) =
x
ln
x
, and thus
by applying the chain rule we have
g
0
(
x
) =
F
0
(
x
2
)
·
2
x

F
0
(2
x
)
·
2 =
x
2
ln(
x
2
)
·
2
x

2
x
ln(2
x
)
·
2 = 4
x
3
ln(
x
)

4
x
ln(2
x
)
,
where we have used ln(
x
2
) = 2 ln
x
for the last identity.
b) We are going to compute the integral
x
2
Z
2
x
t
ln
tdt
by parts. Since we want to get the ln
t
diﬀerentiated, we will use
u
= ln
t
, which
forces us to have
dv
=
tdt
, so that
du
=
dt
t
and
v
=
t
2
2
. We thus obtain
g
(
x
) =
x
2
Z
2
x
t
ln
tdt
=
±
t
2
ln
t
2
²
x
2
2
x

x
2
Z
2
x
t
2
dt
2
t
2
=
±
t
2
ln
t
2
²
x
2
2
x

x
2
Z
2
x
tdt
2
=
±
t
2
ln
t
2

t
2
4
²
x
2
2
x
=
x
4
ln(
x
2
)

4
x
2
ln(2
x
)
2

x
4

4
x
2
4
=
x
4
ln(
x
)

2
x
2
ln(2
x
) +
x
2

x
4
4
,
where we have again used ln(
x
2
) = 2 ln
x
. We now use the chain and product
rules to compute the derivative
g
0
(
x
) = 4
x
3
ln(
x
) +
x
4
x

4
x
ln(2
x
)

2
x
2
·
2
2
x
+ 2
x

x
3
=
4
x
3
ln(
x
) +
x
3

4
x
ln(2
x
)

2
x
+ 2
x

x
3
= 4
x
3
ln(
x
)

4
x
ln(2
x
)
matching the answer in part a).
3
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View Full Document Problem 2
Evaluate the following deﬁnite integrals:
1.
2
Z
0
x
2
p
4

x
2
dx
2.
e
π
Z
1
cos(ln
x
) sin
2
(ln
x
)
x
dx
3.
2
/π
Z
1
/π
sin(1
/x
)
x
2
dx
Solution:
1. In this case a trigonometric substitution is called for. We substitute
x
=
2 sin
θ
, in which case
p
4

x
2
=
p
4

4 sin
2
θ
=
√
4 cos
2
θ
= 2 cos
θ
and
dx
=
d
(sin
θ
) = cos
θdθ
so we get (don’t forget to change the limits of integration, using
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This note was uploaded on 10/21/2010 for the course MAT 126 taught by Professor Sutherland during the Spring '07 term at SUNY Stony Brook.
 Spring '07
 sutherland
 Math, Derivative, Fundamental Theorem Of Calculus

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