Pre-Calc Homework Solutions 341

# Pre-Calc Homework Solutions 341 - Section 8.3 dx x4 dx 0 1...

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Unformatted text preview: Section 8.3 dx x4 dx 0 1 0 341 44. 2 1 1 0 dx x4 dx x4 1 1 by symmetry about the y-axis dx 1 47. For x Area 0, y 0 on [1, ). b 1 ln x dx x2 lim b 1 ln x dx x2 x4 dx x4 1 x4 1 dx x2 1 0 x4 1 exists because 1 1 on [1, x2 b 1 x4 1 Integrate exists on [0, 1]. u du ln x 1 dx x ln x dx by parts. x2 dv v ln x x dx x2 ln x x ln b b 1 x b 0 ). dx x2 1 x ln x x 1 x 1 lim b 1 x x 1 b 2 dx b 1 1 ln x x2 C ln b b 1 b lim b 1 Area 1 lim b lim 1 1 1 b lim b 1 Note that lim b lim b 1/b 1 0. Since this integral converges, the given integral converges. 48. For x 45. Integrate du (1 dy 1 y2 dy y 2)(1 tan du 1 0, y 0 on [1, ). b (1 dy y )(1 tan 2 1 y) by letting u tan 1 y so y) 1 u ln 1 ln 1 0 u tan b 0 C 1 ln x ln x dx lim dx x 1 b 1 x ln x dx Integrate dx by letting u ln x so du . x x ln x 1 2 1 dx u du u C (ln x)2 C x 2 2 b 1 1 Area lim (ln x)2 lim (ln b)2 2 b b 2 1 Area y C 1 49. (a) The integral in Example 1 gives the area of region R. y) (1 dy y 2)(1 tan 1 y) lim b (1 dy y 2)(1 tan Area 1 dx x lim ln 1 b tan tan 1 1 b (b) Refer to Exploration 2 of Section 7.3. y 0) 1 x2 y 0 lim (ln 1 b b ln 1 The integral converges. The surface area of the solid is given by the following integral. 2 1 2 1 x 1 1 2 dx x2 2 1 1 x x4 x4 1 dx 46. 0 e y dy y2 1 0 e y dy y2 1 0 e y dy y2 1 2 Since 0 1 x x4 1 on [1, x3 1 x4 1 dx x3 e y dy diverges since y2 1 e y2 y ), the direct lim y 1 2 y y lim ey 1 ey 2y y lim ey y 2 comparison test shows that the integral for the surface area diverges. The surface area is . (c) Volume 1 lim Thus the given integral diverges. 1 2 dx x 1 1 dx x2 b lim b 1 1 dx x2 1 x 1 b b 1 lim b lim b 1 (d) Gabriel's horn has finite volume so it could only hold a finite amount of paint, but it has infinite surface area so it would require an infinite amount of paint to cover itself. ...
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