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CHAPTER 23
23.29
InJ.
In^lnS'^-lnS
23.30
In 25.
In25 = ln5
2 = 21n5.
23.31
In
23.32
In
23.33
Ini.
In £ = In (20)"
1 = -In 20 = -ln(4 • 5) = -(In 4 + In 5) = -(In 2
2 + In 5) = -(2 In 2 + In 5).
23.34
In 2
12
.
In 2
12 = 12 In 2.
23.35
Find an equation of the tangent line to the curve y — lnx at the point (1,0).
The slope of the tangent line is y' = l/x = l. Hence, a point-slope equation of the tangent line is
y = x — 1.
23.36
Find the area of the region bounded by the curves y = x
2 , y = 1 /x, and x = \ (see Fig. 23-1).
Fig. 23-1
The curves intersect at (1,1). The region is bounded above by v = l/x. Hence, the area is given by
23.37 Find the average value of l/x on [1,4].
The average value
In 2.
23.38 Find the volume of the solid obtained by revolving about the jc-axis the region in the first quadrant under
and x = 1
We use the disk formula: V= TT
23.39
Sketch the graph of y = ln(* + l).
See Fig. 23-2. The graph is that of y = In x, moved one unit to the left.
7r(2ln2)=2irln2.
y
2 dx = IT
= «-(lnl-lni)=»r(0 + ln4) =
y = x
between x - \
dx = IT Inx]\, 4
dx =
l i\nx]*= Kln4-lnl)= |(21n2-0)= \
In
ln2
1/2 =Un2.
In
= ln5
1/3 =iln5.
CHAPTER 23
23.29
InJ.
In^lnS'^-lnS
23.30
In 25.
In25 = ln5
2 = 21n5.
23.31
In
23.32
In
23.33
Ini.
In £ = In (20)"
1 = -In 20 = -ln(4 • 5) = -(In 4 + In 5) = -(In 2
2 + In 5) = -(2 In 2 + In 5).
23.34
In 2
12
.
In 2
12 = 12 In 2.
23.35
Find an equation of the tangent line to the curve y — lnx at the point (1,0).
The slope of the tangent line is y' = l/x = l. Hence, a point-slope equation of the tangent line is
y = x — 1.
23.36
Find the area of the region bounded by the curves y = x
2 , y = 1 /x, and x = \ (see Fig. 23-1).
Fig. 23-1
The curves intersect at (1,1). The region is bounded above by v = l/x. Hence, the area is given by
23.37 Find the average value of l/x on [1,4].
The average value
In 2.
23.38 Find the volume of the solid obtained by revolving about the jc-axis the region in the first quadrant under
and x = 1
We use the disk formula: V= TT
23.39
Sketch the graph of y = ln(* + l).
See Fig. 23-2. The graph is that of y = In x, moved one unit to the left.
7r(2ln2)=2irln2.
y
2 dx = IT
= «-(lnl-lni)=»r(0 + ln4) =
y = x
between x - \
dx = IT Inx]\, 4
dx =
l i\nx]*= Kln4-lnl)= |(21n2-0)= \
In
ln2
1/2 =Un2.
In
= ln5
1/3 =iln5.
