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Unformatted text preview: in which case
graph of g 1 = A(x + 3) + B (x − 1) . 1 To determine A, set x = 1. Then
A= 1
.
4 On the other hand, to determine B , set x =
−3. Then
1
B=− .
4
Thus
I= 1
1
dx −
x−1
4 1
4 1
dx .
x+3 graph of f
−3 3 (axes not drawn to scale). To express the area
as a deﬁnite integral we need to ﬁnd where
the graphs intersect, i.e. when
7
= 1.
16 − x2 Consequently,
I= 1
ln
4 x−1
x+3 +C Thus the graphs intersect when x = ±3.
Hence
3 Area = with C an arbitrary constant.
007 f ( x) = 7
,
16 − x2 1. Area = 3 − g ( x) = 1 . 77
sq.units
ln
82 2. Area = 3 − 7
ln 7 sq.units
4 3. Area = 6 − 7
ln 7 sq.units correct
4 4. Area = 3 + 3 10.0 points Find the bounded area enclosed by the
graphs of f and g when 7
ln 7 sq.units
8 (g (x) − f (x)) dx
−3 =2 1−
0 7
dx
16 − x2 since both f and g are even functions. To
evaluate this integral we must ﬁrst use partial
fractions:
7
7
=
2
16 − x
(4 − x)(4 + x)
= 7
1
1
.
+
8 4−x 4+x Thus the area is gi...
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This homework help was uploaded on 03/21/2014 for the course M 408S taught by Professor Stepp during the Spring '11 term at University of Texas at Austin.
 Spring '11
 STEPP
 Calculus

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