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6x dx Evaluate the indefinite integral: 1 (1-4)2F""(+h Determine if the series converges absolutely; conditionally. or not at all: 0 + $...

Question

6x dx Evaluate the indefinite integral: 1 (1-4)2F""(+h Determine if the series converges absolutely; conditionally. or not at all: 0 + $

6x dx Evaluate the indefinite integral: 1 (1-4) 2F""(+h Determine if the series converges absolutely; conditionally. or not at all: 0 + $



Answers

Determine whether the integral converges or diverges. Find the value of the integral if it converges. $$\int_{0}^{1} \ln x d x$$

All right. In this problem we wish to determine the convergence of the improper integral. The integral from one to infinity of six divided by extra fourth. Yes, this question challenges our understanding of the convergence of improper integral. Remember that an integral is improper if it possesses one or both of these qualities, one wonder both limited the immigrant integration of infinite to there's an infinite discontinuity on the bounds A. B. So we see that are integral is discontinuous because it has bound infinity to evaluate what the convergence with improper integral. We have to determine whether or not the relevant limits exist. That is limited. We approached infinity integral one to be six or X to the fourth. Yet so are valuing this limit. It gives women as he approaches infinity negative two over execute values and want to be. It limits the approach infinity negative two over B. Q. Plus to the first value goes to zero because anything divided by infinity is zero. Thus our limit evaluates as zero plus two equals two. Therefore the limit exists and we say that the integral converges to value to

All right. In this problem we wanted to share one of the given integral converges and if it does compute its value, the integral from zero to infinity of X squared E to the negative six X dx. This question is challenging an understanding of the topic of improper integral evaluation and proper and the girls are said to converge when the appropriate limits converged. Either in the case of an infinite limit of integration and or this case secondly of an infinite is concentrated from the bounds of the limits of integration. So proceeding to solve we said we evaluate the integral as convergent if the appropriate. LTD said to converge as well limit as N approaches infinity integral zero to explore the negative six X. Dx. So thus we take a limited A approaches infinity of negative 18 expert for six X plus one over 100 and 86 X which is the anti derivative, plugging in the bound zero to A by the fundamental families calculus. This was taken by integration by parts. This completes the limited approaches infinity on the right because either the six A is a syntactical greater than any polynomial. This goes to zero. So the limit does converge does integral to zero plus one over 190

For this problem we need to determine if the improper integral converges or diverges. Now the immigrant is discontinuous at x equals zero and the integral is improper because ah the upper limit is infinite. So so we need to a approaches zero from the right of the integral from a two one of four over square at the vax times X plus six the X plus we have limits as B approaches infinity of the integral from one to be of four over square root of X times X plus six dx. And then from here we want to apply substitution. We let U equal to square the vex and we get U squared equals X. So that two U. D. U. is DX. And so from here we have Limit as a approaches zero from the right of the integral from 8 to 1 of for over you times U squared plus six times to you. Do you? Plus limit as B approaches infinity of the integral from one to be a four over you Times you squared plus six times to you, do you. And in here we can get rid of the U. And we have Limit as a approaches zero from the right of the integral. From 81 of eight over U squared plus six D. U plus limit as we approaches infinity of the integral from one to be of eight. Overuse squared plus six D. You now note that the integral of one over U squared plus a squared d'you? This is equal to one over a tangent inverse of you over A. And so we have limits As he approaches zero from the right of he has eight over the squares of six tangent inverse of You over the squares of six. This evaluated from 8 to 1 plus limits as we approaches infinity of eight over squared of six tangent inverse of You over squared the six. This evaluated from one to be Now, since you square the vex, we have limit as they approach is zero from the right of eight over square to six tangent inverse of squared of X over the square root of six. Or that's just the squared of X over six. This evaluated from 81 plus limits as B approaches infinity of eight over the squares of six tangent inverse. Ah, square it of X over six, Evaluated from 1 to be evaluating Now the integral we have limits As they approach is zero from the right of eight squared of 60. Ancient inverse of one over the square root of six -8 over the square root of six. The engines in verse. Uh, Square it of 8/6. And then plus we have limits as B approaches infinity of eight over squared of six tangent inverse of We have square it of be over six -8 over the square root of six tangent inverse. We have square with of 1/6. And then this will cancel out. And at a approaching zero This will become zero. And so we are left with this term, which is equal to eight over the squares of six tangent inverse of infinity, Which is just eight Over the square root of six times pi over two, or that's equal to For by over the square root of six. And so this is a finite value for the improper integral, which tells us that the improper integral converges.

In the problem, since one of the integration is minus infinity. Therefore we have the integral is an improper integral. Now This is written as integration minus infinity to zero except on X squared plus one day X. So this is written as a limit attending to minus infinity one upon to integration 8-0 DT upon T Or we have excess choir plus one SD. This is twice of X dx. Nettie poster DT. This is written as limit, attending to minus infinity half limit model T 8-0, Softer putting the limits, we have this as infinity. Therefore it never says this is the answer.


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