PostClass 7.4-solutions

# 1 to determine a set x 1 then a 1 4 on the other

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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.

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