Math1B Quiz3Solutions

# Math1B Quiz3Solutions - p then 1 (1 + x 2 ) p ≤ 1 x 2 p ....

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QUIZ 3 SOLUTION Question 1 (1) The integral is improper because (1) the interval of integration is inﬁnite (2) the integrand has a discontinuity at x = 0 so we need to split the integral in two parts Z 0 1 x p dx = 1 Z 0 1 x p dx + Z 1 1 x p dx, The ﬁrst integral converges when p < 1 and the second converges when p > 1 that means that the total integral will never converge! (2) Observe that the integral is improper only because the interval of integration is inﬁnite. Then we can write it in the following manner Z 0 1 (1 + x 2 ) p dx = 1 Z 0 1 (1 + x 2 ) p dx + Z 1 1 (1 + x 2 ) p dx, So the ﬁrst integral on the right hand side is a regular integral which always converges. We apply comparison theorem to the second one. Clearly 1 + x 2 x 2 Therefore 1 1 + x 2 1 x 2 . Raise it to the power of p for a positive

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Unformatted text preview: p then 1 (1 + x 2 ) p ≤ 1 x 2 p . This means that the integral ∞ R 1 1 (1+ x 2 ) p dx converges when the integral ∞ R 1 1 x 2 p dx which hap-pens when 2 p > 1. To show that it doesn’t converge for other values, we will use comparison theorem again. This time we can write 1 + x 2 ≤ 2 x 2 , since ,x ≥ 1 . Then doing similar manipulations we arrive at the conclusion that 1 (1 + x 2 ) p ≥ 1 2 p x 2 p . Therefore the integral ∞ R 1 1 (1+ x 2 ) p dx diverges whenever the integral ∞ R 1 1 2 p x 2 p dx = 1 2 p ∞ R 1 1 x 2 p dx diverges. Which happens for p ≤ 1 / 2. 1 2 Question 2 See your lecture notes from 2/5....
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## This note was uploaded on 07/10/2008 for the course MATH 1B taught by Professor Reshetiken during the Spring '08 term at Berkeley.

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Math1B Quiz3Solutions - p then 1 (1 + x 2 ) p ≤ 1 x 2 p ....

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