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Unformatted text preview: 216 Section 5.3
2 3 19. Divide the shaded area as follows.
y 2.25 R1 R2 4 x y = 3x x 2 (b) Area
0 (x 2 6x F(0)] 0 8) dx
2 (x 2 F(2)] 6x 8) dx [F(2)
20 3 [F(3)
20 3 6 22 3 22. 4 Note that an antiderivative of 3x
3 x 2 is F(x) F(0) Area of R1 Area of R2 (3x
0 4 x 2) dx x 2) dx F(3)]
9 2 9 2 11 6 11 6 F(3) 3 2 1 3 x x . 2 3 9 9 0 2 2 [0, 2] by [ 5, 3] An antiderivative of F(x)
2 x2 4x. 4) dx 5x F(0)] 0 5x 4 is (3x
3 1 3 x 3 5 2 x 2 [F(4)
8 3 (a)
0 ( x2 5x
1 F(2) 4) dx F(0)
2 1 2 3 0 5x 2 3 Total shaded area 19 3 (b) Area
0 ( x2 ( x2 4) dx [F(1) 20. Divide the shaded area as follows.
y 3 y= x2 2x [F(2)
2 3 F(1)]
11 6 11 6 3 23. R2 R1 3 x [0, 3] by [ 3, 2] An antiderivative of 2x 2x is F(x)
1 3 x 3 x 2 is F(x) F(0)
3 x2 0 0 1 3 x . 3 Note that an antiderivative of x 2
2 x 2. 3 (a)
0 (2x x 2 ) dx
2 F(3) x 2 ) dx F(0)] 0 Area of R1 (x 2 2x) dx (b) Area F(0)] 0 2x) dx F(2)
4 3 4 3 4 3 8 3 4 3 0 (2x
0 (2x
2 x 2) dx F(2)] [F(2)
4 3 [F(2)
4 3 [F(3)
4 3 8 3 3 0 0 Area of R2 (x 2 2 F(3) 0 Total shaded area 21.
4 3 24. [0, 5] by [ 5, 5] An antiderivative of x 2
5 4x is F(x) F(0) 1 3 x 3 25 3 2x 2. 0
25 3 (a)
[0, 3] by [ 1, 8]
0 (x 2 4x) dx F(5) An antiderivative of x 2
3 6x F(3) 8 is F(x) F(0) 6 1 3 x 3 3x 2 6 8x. (a)
0 (x 2 6x 8) dx 0 ...
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 Spring '08
 GERMAN
 Derivative

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