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19. Divide the shaded area as follows. Note that an antiderivative of 3 x 2 x 2 is F ( x ) 5 } 3 2 } x 2 2 } 1 3 } x 3 . Area of R 1 5 E 3 0 (3 x 2 x 2 ) dx 5 F (3) 2 F (0) 5 } 9 2 } 2 0 5 } 9 2 } Area of R 2 5 2 E 4 3 (3 x 2 x 2 ) dx 5 2 [ F (4) 2 F (3)] 5 2 1 } 8 3 } 2 } 9 2 } 2 5 } 1 6 1 } Total shaded area 5 } 9 2 } 1 } 1 6 1 } 5 } 1 3 9 } 20. Divide the shaded area as follows. Note that an antiderivative of x 2 2 2 x is F ( x ) 5 } 1 3 } x 3 2 x 2 . Area of R 1 5 2 E 2 0 ( x 2 2 2 x ) dx 5 2 [ F (2) 2 F (0)] 5 2 1 2 } 4 3 } 2 0 2 5 } 4 3 } Area of R 2 5 E 3 2 ( x 2 2 2 x ) dx 5 F (3) 2 F (2) 5 0 2 1 2 } 4 3 } 2 5 } 4 3 } Total shaded area 5 } 4 3 } 1 } 4 3 } 5 } 8 3 } 21. [0, 3] by [ 2 1, 8] An antiderivative of x 2 2 6 x 1 8 is F ( x ) 5 } 1 3 } x 3 2 3 x 2 1 8 x . (a) E 3 0 ( x 2 2 6 x 1 8) dx 5 F (3) 2 F (0) 5 6 2 0 5 6 (b) Area 5 E 2 0 ( x 2 2 6 x 1 8) dx 2 E 3 2 ( x 2 2 6 x 1 8) dx 5 [ F (2) 2 F (0)] 2 [ F (3) 2 F (2)] 5 1 } 2 3 0 } 2 0 2 2 1 6 2 } 2 3 0 } 2 5 } 2 3 2 } 22. [0, 2] by [ 2 5, 3] An antiderivative of 2 x 2 1 5 x 2 4 is F ( x ) 5 2 } 1 3 } x 3 1 } 5 2 } x 2 2 4 x . (a) E 2 0 ( 2 x 2 1 5 x 2 4) dx 5 F (2) 2 F (0) 5 2 } 2 3 } 2 0 5 2 } 2 3 } (b) Area 5 2 E 1 0 ( 2 x 2 1 5 x 2 4) dx 1 E 2 1 ( 2 x 2 1 5 x 2 4) dx 5 2 [ F (1) 2 F (0)] 1 [ F (2) 2 F (1)] 5 2 1 2 } 1 6 1 } 2 0 2 1 3 2 } 2 3 } 2 1 2 } 1 6 1 } 24 5 3 23. [0, 3] by [ 2 3, 2] An antiderivative of 2 x 2 x 2 is F ( x ) 5 x 2 2 } 1 3 } x 3 . (a) E 3 0 (2 x 2 x 2 ) dx 5 F (3) 2 F (0) 5 0 2 0 5 0 (b) Area 5 E 2 0 (2 x 2 x 2 ) dx 2 E 3 2 (2 x 2 x 2 ) dx 5 [ F (2) 2 F (0)] 2 [ F (3) 2 F (2)] 5 1 } 4 3 } 2 0 2 2 1 0 2 } 4 3 } 2 5 } 8 3 } 24. [0, 5] by [ 2 5, 5] An antiderivative of x 2 2 4 x is F ( x ) 5 } 1 3 } x 3 2 2 x 2 . (a) E 5 0 ( x 2 2 4 x ) dx 5 F (5) 2 F (0) 5 2 } 2 3 5 } 2 0 5 2 } 2 3 5 } y 3 x 3 R 1 R 2 y = x 2 – 2 x y 2.25 –4 x 4 R 1 R 2 y = 3 x x 2 216 Section 5.3

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(b) Area 5 2 E 4 0 ( x 2 2 4 x ) dx 1 E 5 4 ( x 2 2 4 x ) dx 5 2 [ F (4) 2 F (0)] 1 [ F (5) 2 F (4)] 5 2 1 2 } 3 3 2 } 2 0 2 1 3 2 } 2 3 5 } 2 1 2 } 3 3 2 } 24 5 13 25. An antiderivative of x 2 2 1 is F ( x ) 5 } 1 3 } x 3 2 x . av 5 } ˇ 1 3 w } E ˇ 3 w 0 ( x 2 2 1) dx 5 } ˇ 1 3 w } 3 F ( ˇ 3 w ) 2 F (0) 4 5 } ˇ 1 3 w } (0 2 0) 5 0 Find x 5 c in [0, ˇ 3 w ] such that c 2 2 1 5 0 c 2 5 1 c 5 6 1 Since 1 is in [0, ˇ 3 w ], x 5 1. 26. An antiderivative of 2 } x 2 2 } is F ( x ) 5 2 } x 6 3 } . av 5 } 1 3 } E 3 0 1 2 } x 2 2 } 2 dx 5 } 1 3 } [ F (3) 2 F (0)] 5 } 1 3 } 1 2 } 9 2 } 2 5 2 } 3 2 } Find x 5 c in [0, 3] such that 2 } c 2 2 } 5 2 } 3 2 } . c 2 5 3 c 5 6 ˇ 3 w Since ˇ 3 w is in [0, 3], x 5 ˇ 3 w . 27. An antiderivative of 2 3 x 2 2 1 is F ( x ) 5 2 x 3 2 x . av 5 } 1 1 } E 1 0 ( 2 3 x 2 2 1) dx 5 F (1) 2 F (0) 5 2 2 Find x 5 c in [0, 1] such that 2 3 c 2 2 1 5 2 2 2 3 c 2 5 2 1 c 2 5 } 1 3 } c 5 6 } ˇ 1 3 w } Since } ˇ 1 3 w } is in [0, 1]. x 5 } ˇ 1 3 w } .
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