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1/2 pointsPrevious AnswersSESSCALCET2 6.6.018.Deterine whether the integral is convergent or divergent:rer/4 drconvergentdivergentIfit is convergent, evaluate it: (...

Question

1/2 pointsPrevious AnswersSESSCALCET2 6.6.018.Deterine whether the integral is convergent or divergent:rer/4 drconvergentdivergentIfit is convergent, evaluate it: (If the quantity diverges enter DIVERGES.)Need Help?Reed ItIlelu TutorSubmit AnswerPractice Another Version

1/2 points Previous Answers SESSCALCET2 6.6.018. Deterine whether the integral is convergent or divergent: rer/4 dr convergent divergent Ifit is convergent, evaluate it: (If the quantity diverges enter DIVERGES.) Need Help? Reed It Ilelu Tutor Submit Answer Practice Another Version



Answers

Evaluate the integrals. If the integral diverges, answer "diverges."
$$
\int_{-\infty}^{0} \frac{d x}{x^{2}+1}
$$

Yeah. In this problem we wish to determine the convergence of the improper integral. The integral from negative infinity to negative 10 of D X over X squared. This question challenges our understanding of integration techniques related to improper integral. Before we evaluate, let's remind ourselves of improper integral. Czar intervals are improperly possess either of two qualities. One wonderful limits of integration is infinite, that's true original and two there's an infinite continuing on A to B. Are integral has discontinuity on zero which is not inside our limits. So we only refer to reason one to evaluate the convergence of an integral such as this. We say that it converges at the relevant limits exist for us. That's a limited approaches negative infinity from eight negative 10 D X over X square. Thus let's evaluate the limits. Us even exists. The women and the approaches infinity of the integral is equal to the limit. The approach is very negative. One over x 89 to 10. This limited the approaches infinity 1/10 plus one over a. The second term goes to zero because divided by infinity goes to zero. That's what left with 1/10. Our limit exists, which means are integral converges to 1/10.

We're looking at the improper integral going from two to infinity of one over X squared. Let's go ahead and rewrite this and now we can use We can undo power rule to find the anti derivative and as we approach infinity, notice that this evaluates to zero. So let's gives us one half so does converge on a convergence toe one half.

Here we have the integral from zero to infinity of DX X plus one to the three halves. We'll start by rewriting it X plus one to the negative. Three has DX and now we confined the anti derivative negative two x plus one to the negative one half. Now we can write this thing down as a limit. His X goes to infinity negative to to the X Plus one the negative one half plus two tons. One The first term goes to zero So we just have to, particularly those things convergence.

In the problem we have been given integration one to infinity since one of the limit is infinitely. Therefore this integration is an improper integral and we have to found Now we will be evaluating it to get whether it convinces or not. So this is written as limit be tending to infinity doesn't want to be f x dx that equals to integration or rather limited be tending to infinity digression, want to be experiments to dx. This equals to limit be tending to infinity. Actually Power -1 upon -1. Here it is one to be And this equals two, limit beaten into infinity minus one upon X putting the limits one and be Therefore the z equals two -1 upon infinity plus one. This is zero and city is equal to one, therefore it is converted. Hence this is the answer.


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