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Evaluate the following integral or state that it diverges_dxx +441Select the correct choice below and, if necessary; fill in the answer box to complete your choiced...

Question

Evaluate the following integral or state that it diverges_dxx +441Select the correct choice below and, if necessary; fill in the answer box to complete your choicedxOA(Type an exact answer; using T as needed:)x2+4410 B. The improper integral diverges_

Evaluate the following integral or state that it diverges_ dx x +441 Select the correct choice below and, if necessary; fill in the answer box to complete your choice dx OA (Type an exact answer; using T as needed:) x2+441 0 B. The improper integral diverges_



Answers

Evaluate the integrals. If the integral diverges, answer "diverges."
$$
\int_{0}^{e} \ln (x) d x
$$

So in this costume, first rewrite, the integral is the limit. Let's be goes to infinity. These you're eating the power of native actually Axe. And now we're gonna integrate, and we're gonna apply that even the power of negative a ax. The ax is you put a negative one of Ray either The power of negative X plus our constant See. So we end up with this. When we integrate, we end up with the negative limited. Just be goes to infinity of either the power negative eyes e j. So now we can use the fundamental theory of calculus from the hand of Got anything negative limit as he goes to infinity. You the power of negative B minus one. So when we evaluate this limit, this is equal to negative. Then it is the power of negative entity minus one, which is negative zero minus one, which is one that's the answer to this question.

To evaluate this improper integral. We know that by definition this is equal to the limit is be approaches infinity of the integral from one to be of X times erased. Negative two X. Dx. And then from here we will apply integration by parts and here we will let U equal to X. And DV equal to erase the negative two X. Dx. So that differential of U. Is equal to the differential of X. And V. Which is the anti derivative of devi is equal to erase the negative two. X over negative two. So by integration by parts we have the integral of X times erased negative two X. Dx. This is equal to U. Times V. That's negative X times erase the negative two. X over two minus the integral of E times do you? That's erase the negative two X over negative two dx. And that's the same as negative X times erase the negative two X over two plus one half times erase the negative two X over negative two. Or that's the same as negative X times erase the negative two X over two minus, erase the negative two X over four. Now we can combine this into the negative of two x plus one over four times erase 2 2 x. So the integral becomes limit. As the approaches infinity of the negative of two X plus 1/4 Times erase The two X. This evaluated from 1 to be. Now if X is equal to be then we have this equal to the limit is be approaches infinity of negative two times B plus 1/4 times erased. To be minus negative. That will be plus two times one plus one. That's 3/4 times erased too. And if you take the limit we know that this will give us infinity over infinity. So we have to apply loopholes rule and from there we will get limit as B approaches negative or positive infinity times the negative of the derivative of the numerator, which is to over four Times two times erased. It to be Plus 3/4 Times erased to the 2nd power. Now if you plug in infinity this will give us -2 over infinity plus 3/4 times erase the second power and this goes to zero. So we have 3/4 times E squared. And since this is finite, we say that the improper integral converges.

To evaluate the given improper integral. We know that by definition this is equal to the limit. As a approaches negative infinity. Are the integral from 80 of x squared times erase two X dx. Now from here we will use tabular integration and he didn't want to let F of X equal to X squared. And our G of X is equal to the race to X. And then we will form a table for the derivatives of F. Of X. And then we have the sign or operation and then we have G of X in its into growth. So we start with X squared and a positive operation or sign. And then you have G. Fx which is erased two X. Now we do derivatives. First we want to take the derivative of X square. That will give us two X. And then we take the derivative of two X. We get to and then we get the derivative of two of this will give us zero. And then from here we take the integral of G of X. That will be itself. Since this is erased. Two X. So we have erased two X, erase two X. And then erased X. And then the operation will be alternating. So from plus we have minus and then plus and then minus. Next we want to multiply elements from F. And elements from G in diagonal format. So the products will be X squared erased X. We have minus two x, erase two x. and then plus we have to times erase two X. So by tabular integration we have the integral of X squared times erase two X dx. This is equal to x squared times erase x minus two X times eraser X plus two times eraser X plus C. Our that's the same as E res two X times X squared minus two X plus two. And then plus C. So by fundamental theorem of calculus than we have limit as he approaches negative infinity. Uh The race to X times X squared minus two X plus two. That's evaluated from +80 When X0. We have limits as a approaches negative infinity of erased zero times 0 zero plus two minus when X is a. We have erased to eight times a squared minus to a plus two. And from here we will get 2- the limit as a approaches negative infinity. Ah A squared minus to a plus two over erase the negative aid. And this will be infinity over infinity. So we have to apply Lou Beatles rule and we will get to minus the limit as a approaches negative infinity of the derivative of the numerator. That's two A -2 over negative erased negative a which is still infinity over infinity. So we have to apply the rule again. And this will give us two minus the limit as a approaches negative infinity of the derivative of the Human radar. That's two over the race to negative A. And if we apply Or evaluate the limit we get to -2 over erase infinity, Which goes to zero. So this will give us a value equal to two. And because this is finite, we say that the improper integral converges.

To evaluate this improper integral. We know that by definition this is equal to the limit as the approaches infinity of the integral from zero to be of X plus one times erase to negative X. Dx. Now from here we have to apply integration by parts. We want to let U equal to X plus one and D. V is equal to erase the negative X. Dx. So that the differential of U. Is equal to the differential effects and V is equal to negative erased and negative X. So by integration by parts The integral of X-plus one times erase the negative X. Dx is equal to U times V. That's negative of express one times erase the negative X minus the integral of V times D. You that's negative, erase the negative X. D. X. And if we integrate this we have negative of X plus one times erase the negative X plus we have negative erase to negative X. Or that the same as the negative of erase the negative X times X plus one plus one that's X plus two. Are we can write it as negative of explodes to over erase two X. Now by fundamental theorem of calculus that is equal to the limit to be approaches infinity uh negative of X plus two over erase two X. This evaluated from zero to be. So if X is B we have LTD be approaches infinity of negative B plus two over erase to be minus negatives or plus when X zero that'll be zero plus two or two over erased. Zero. And if we plug in infinity we get here infinity over infinity. So we have to apply loop. It does rule and this is equal to the limit as be a purchase infinity. The negative of the derivative of B plus two. That's one over a race to be And then plus two. This is equal to negative one over erase infinity plus two, and we know that this goes to zero. So we have a value equal to two which is finite, meaning that the improper integral converges.


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