MATH 138
Assignment 1solution
Due by:
11 a.m.
, Friday, January 15.
1. This exercise is intended to review some aspects of the fundamental theorem of calculus, the
increasing function theorem, as well as inverse functions. The calculations are not hard, but
you need to keep the concepts straight in your head.
(a) Prove that the integral function
g
(
x
) =
x
−
1
√
1
−
t
2
dt
is increasing on the interval
[
−
1
,
1]
. This takes very little writing.
(b) Sketch the graph of
g
over
[
−
1
,
1]
.
Hint. This can be done by thinking of
g
(
x
)
as the area from
−
1
to
x
under the semi
circle
y
=
√
1
−
t
2
. There is no need to explicitly calculate
g
(
x
)
.
(c) What is the range of
[
−
1
,
1]
under the function
g
?
Again, think of area under
y
=
√
1
−
t
2
.
(d) Explain brie
fl
y why
g
(0) =
π/
4
.
If
h
is the inverse function of
g
,
fi
nd
h
(
π/
4)
.
Solution:
(a) By the fundamental theorem of calculus, we have
g
(
x
) =
√
1
−
x
2
, which is
≥
0
for
x
∈
[
−
1
,
1]
. Thus
g
(
x
)
is increasing.
(b) Since
g
(
x
) =
√
1
−
x
2
, we have
g
(
x
) =
1
2
(1
−
x
2
)
−
1
/
2
(
−
2
x
) =
−
x
√
1
−
x
2
.
Thus
g
(
x
)
is increasing, and is concave upward on
[
−
1
,
0]
and downward on
[0
,
1]
.
x
y
y
=
g
(
x
)
1
−
1
(c) Notice that
g
(
−
1) = 0
. Also, since
g
(1)
represents the area of a semicircle of radius 1,
we have
g
(1) =
π/
2
. Since
g
(
x
)
is continuous and increasing, the range of
g
is
[0
, π/
2]
.
(d) Since
g
(0)
represents
1
/
4
of the the area of a circle of radius 1, we have
g
(0) =
π/
4
.
Also, if
h
(
g
(
x
)) =
x
, then
h
(
g
(
x
))
g
(
x
) = 1
. By putting
x
= 0
and recalling
g
(
x
) =
√
1
−
x
2
, we have
h
(
g
(0))
g
(0) = 1
=
⇒
h
(
π/
4) =
1
g
(0)
=
1
√
1
−
0
2
= 1
.
1
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2. Find the following inde
fi
nite integrals. A suitable substitution should work in each case.
(a)
1
x
ln
x
dx,
(b)
arctan
x
1 +
x
2
dx,
(c)
sin
5
x
cos
x dx,
(d)
dx
1 + sin
x
.
Hint for (d). Multiply the top and bottom by
1
−
sin
x
, use a trigonometric identity, and then
make a substitution.
Solution:
(a) Let
u
= ln
x
and
du
=
1
x
dx
. Then
1
x
ln
x
dx
=
1
d
du
= ln

u

+
C
= ln

ln
x

+
C.
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 Winter '07
 Anoymous
 Fundamental Theorem Of Calculus, Inverse Functions, Inverse function, dx

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