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Pre-Calc Homework Solutions 216

# Pre-Calc Homework Solutions 216 - 216 Section 5.3 2 3 19...

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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
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