Team Homework 3.docx

# Because 9 x therefore 4 x 2 1 1 2 9 x lim b 1 2 b 1

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Because 9 x , therefore 4 x 2 1 ¿ 1 / 2 ¿ ¿ 9 x ¿ lim b→ 1 / 2 b 1 ¿ This limit exists and converges, so that means our given integral converges as well. b. 2 9 x 2 + x x 3 i. To determine whether this integral converges or diverges, we have to decide what comparison test we are going to use. For this integral, we are going to use the direct comparison test. So, we find a function that is greater than or less than the given. 9 x 2 + x x 3 x 1 / 2 > 9 x 2 x 3 lim b→ ∞ 2 b 9 x 2 + x x 3 This integral diverges, so we can say that by direct comparison the given integral also diverges. c. 9 x sin ( x )+ 2 x x 2 + cos ( x ) dx i. To determine whether this integral converges or diverges, we have to decide what comparison test we are going to use. For this integral, we are going to use the direct comparison test. First, we have to find a function

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that is easily compared to this function. 2 x > xsin ( x ) , x 2 > cos ( x ) , xsin ( x ) x 2 < x sin ( x )+ 2 x x 2 + cos ( x ) Since this function is greater than our given function, if it’s integral diverges then the given integral diverges.
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• Fall '07
• Irena
• lim, #, 3 M, LCT, 116 Sec

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