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Math 1271
Solutions for Fall 2005 Final Exam
1)
Since the equation
x
2
+
y
2
=
e
xy
cannot be rearranged algebraically in order to
write
y
as an explicit function of
x
, we must instead differentiate this relation
implicitly with respect to
x
, in order to find an expression for
y
’ :
d
dx
(
x
2
+
y
2
)
=
d
dx
(
e
xy
)
⇒
2
x
+
2
y
⋅
dy
dx
=
d
du
(
e
u
)
⋅
du
dx
setting
u
=
xy
⇒
2
x
+
2
y
⋅
dy
dx
=
e
u
⋅
d
dx
(
xy
)
⇒
2
x
+
2
y
⋅
dy
dx
=
e
u
⋅
(
x
'
⋅
y
+
x
⋅
y
' )
⇒
2
x
+
2
y
⋅
dy
dx
=
e
xy
⋅
(1
⋅
y
+
x
⋅
dy
dx
)
⇒
2
y
⋅
dy
dx
−
x
⋅
e
xy
⋅
dy
dx
=
y
⋅
e
xy
−
2
x
⇒
dy
dx
=
y
⋅
e
xy
−
2
x
2
y
−
x
⋅
e
xy
.
(A)
2)
The two parts of the Fundamental Theorem of (Integral) Calculus can be put together
to state that, for a function
h
(
x
)
continuous on
[
a
,
b
] ,
the definite integral
,
with
a
<
x
<
b
,
equals
f
(
x
)
−
f
(
a
) , where
f
’(
x
) =
h
(
x
) .
If we modify the
upper limit of the integration to make it a function
u
(
x
)
in the same interval,
we then
have
The derivative of this integral is then
f
(
a
) is a constant
The integrand given in the problem is continuous everywhere (since it is defined
everywhere), so the Fundamental Theorem and its consequences can be applied here.
For the integral function defined by
sin(
e
t
)
t
2
+
1
x
2
10
∫
dt
,
we must first make an alteration
to accommodate the conditions for the Theorem:
we move the constant limit of
integration to become the lower limit by “reversing the direction” of integration:
h
(
t
)
dt
a
x
"
h
(
t
)
dt
a
u
(
x
)
"
=
f
(
u
(
x
) )
#
f
(
a
)
.
d
dx
h
(
t
)
dt
a
u
(
x
)
"
=
d
dx
f
(
u
(
x
) )
#
f
(
a
)
[ ] =
d
dx
f
(
u
(
x
) )
#
d
dx
f
(
a
)
=
d
du
f
(
u
)
"
du
dx
#
0
=
df
dx
u
(
x
)
"
du
dx
or
h
(
u
(
x
) )
"
u
' (
x
)
.
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View Full Documentsin(
e
t
)
t
2
+
1
x
2
10
∫
dt
=
−
sin(
e
t
)
t
2
+
1
10
x
2
∫
dt
.
With
h
(
x
)
=
sin(
e
x
)
x
2
+
1
and
u
(
x
)
=
x
2
,
the derivative of our integral function is
d
dx
sin(
e
t
)
t
2
+
1
x
2
10
∫
dt
=
−
h
(
u
(
x
) )
⋅
u
' (
x
)
=
−
sin(
e
x
2
)
(
x
2
)
2
+
1
⋅
(
x
2
)'
=
−
2
x
⋅
sin(
e
x
2
)
x
4
+
1
.
(C)
3)
There are many possible ways to choose bounds for the value of a definite integral;
in view of the choices given, which involve integer results, we will want to look at
something fairly simple to calculate.
With the interval being
[ 1 , 2 ] , the smallest value
the integrand itself,
x
4
+
9 sin
2
x
, could
possibly
have would be if the second term
within the radical were zero, reducing the function to
x
4
=
x
2
, which is equal to
1
at
x
= 1 (in fact, the “sinesquared” term is
not
zero there).
With the width of the
interval of integration being
Δ
x
=
2
−
1
=
1 , we can definitely say that
x
4
+
9 sin
2
x
1
2
∫
dx
≥
1
⋅ Δ
x
=
1
.
At the other extreme, the largest value the
integrand could possibly have would be when
sin
2
x
is as large as possible, making the
function equal to
x
4
+
9
⋅
1
=
x
4
+
9
; this takes on its largest value on the interval
at
x
= 2 , where
x
4
+
9
=
4
+
9
=
16
+
9
=
5
.
This result provides an upper
bound for our integral of
x
4
+
9 sin
2
x
1
2
∫
dx
≤
5
⋅ Δ
x
=
5
.
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 Fall '08
 MING
 Calculus, Algebra

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