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(To indicate convergence or divergence, enter one of the words converges or diverges in the appropriate answer blanks.)
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R 1 x 3 + 1 x 4 + 5 x + 3 dx : a similar integrand is so we predict the integral R 1 x x 2 + 5 x + 3 dx : a similar integrand is so we predict the integral R 1 x + 3 x 4 + 6 x 3 + 3 dx : a similar integrand is so we predict the integral R 1 x 2 + 3 x 4 + 5 x 2 + 6 dx : a similar integrand is so we predict the integral Solution: SOLUTION R 1 x 3 + 1 x 4 + 5 x + 3 dx : a similar integrand is 1 x so we predict the integral diverges R 1 x x 2 + 5 x + 3 dx : a similar integrand is 1 x so we predict the integral diverges R 1 x + 3 x 4 + 6 x 3 + 3 dx : a similar integrand is 1 x 3 so we predict the integral converges R 1 x 2 + 3 x 4 + 5 x 2 + 6 dx : a similar integrand is 1 x 2 so we predict the integral converges Answer(s) submitted: 1/x diverges 1/x diverges 1/xˆ3 converges 1/xˆ2 converges (correct) Correct Answers: 1/x diverges 1/x diverges 1/(xˆ3) converges 1/(xˆ2) converges 29. (1 point) Solution: SOLUTION The convergence or divergence of an improper integral de- pends on the long-term behavior of the integrand, not on its short-term behavior. The figure suggests that g ( x ) > f ( x ) for x > k , for some value k . We know that R k f ( x ) dx converges, but because g ( x ) > f ( x ) , this gives us no information about R k g ( x ) dx . Thus we expect that we cannot determine the con- vergence of the latter integral. However, we are interested in R a g ( x ) dx . Breaking the in- tegral into two parts enables us to use what we know about R k g ( x ) dx : Z a g ( x ) dx = Z k a g ( x ) dx + Z k g ( x ) dx . The first integral is finite because the interval from a to k is finite. However, because we don’t know the convergence of R k g ( x ) dx , we remain unable to speak to the convergence of the integral R a g ( x ) dx . Answer(s) submitted: B (correct) Correct Answers: B

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