determine-whether-each-integral-is-convergent-or-divergent-evaluate-those-that-are-convergent-int_-4

Question

Determine whether each integral is convergent or divergent. Evaluate those that are convergent.
$$\int_{-\infty}^{\infty} x e^{-x^{2}} d x$$

Amanda Stone · Accepted Answer

in this problem were given the improper integral X times each the negative X squared DX from negative infinity to infinity and were asked to find whether this is convergent or divergent. So since we have two bounds that go to infinity, I&#x27;m going to rewrite this as two inch girls breaking up that bound. So one is going to have the bounds Negative infinity to zero of our function plus the bounds zero to infinity of our function. So because we have now to and girls ah, with the same terms I&#x27;m going to begin by just solving the indefinite integral X times eats the negative x squared d X that we can use it for both of our integral. So, in order to solve this, we&#x27;re gonna use U substitution where you equals negative X squared and d u equals negative two x d x. So if we plug this in two are integral, we get a negative 1/2 on the side and then we have the integral e to the u D year, and this will just equal 1/2 each the you pussy. If we steps to back in our value for you, that gives us negative 1/2 e to the negative X squared plus c. And finally, uh, we&#x27;re gonna put this into our two intervals. And instead of writing those infinities, I&#x27;m going to use the term T for our first in a girl. I&#x27;m gonna have t go to negative infinity in 1st 2nd in a girl. I&#x27;m gonna have a go to infinity, and this is going to give us negative 1/2 e to the negative x squared from t 20 plus negative 1/2 e to the negative X squared from zero to T where in our first term to goes negative infinity and in our second t goes to infinity. So from here, I&#x27;m gonna take the limit of both of these for a first term, the limit is going to be as t approaches negative infinity. And if we plug in our bounds, we get negative 1/2 plus 1/2 each the negative t squared, plus the limits as t approaches infinity of negative 1/2 e to the negative cheese squared plus 1/2. So if we plug in our limit, both of these terms are going to go to zero because we get e. It&#x27;s the negative infinity for both. And if we look at a really simple sketch of each, the X will notice that as e approaches native infinity, it approaches zero. So this leaves us from both of our terms. Negative 1/2 minus 1/2. Which leaves us with negative 1/4. Oh, I&#x27;m sorry. I got my negatives off here. This leaves us with negative 1/2 plus 1/2 which gives us zero. And in both cases, this gives us a convergent integral.

Amanda Stone · Answer

in this problem were given the improper, integral, X squared times each the negative X cubed D X from negative infinity to infinity and were asked to find whether this is convergent or divergent. So the first thing that we&#x27;re gonna do is we&#x27;re gonna break this up into two inter girls that we don&#x27;t have infinity in both bounds. So I&#x27;m going to such one from negative infinity to zero of our function. I&#x27;m going to set the other from zero to infinity and I&#x27;m gonna add the&#x27;s together. So now we have to integral. And since they have the same function, I&#x27;m going to begin by just solving be indefinite, integral and then plugging in the balance leader. So if we do this, we&#x27;re going to use U substitution to solve this Where you who equal negative x cubed and do you equal negative three X squared D X. And if we plug this end, we&#x27;re gonna get a negative 1/3 up front and then for integral will just have e to you, do you? If we take the integral of this, we just get negative 1/3 each, the U Plus c. And if we substitute back in, Are you value? We get negative 1/3 each, the negative X third plus c. And finally, we&#x27;re gonna go ahead and at our bounds back in. Ah, for first term, I&#x27;m going to have tea approach Negative, infinity. And for a second, I&#x27;m gonna have t approach infinity. So this will give us negative 1/3 each. The negative x cute from t 20 with t equally negative. Infinity plus negative 1/3 e to the negative X cubed from zero to t with t equaling infinity. So from here, we&#x27;re gonna take the limit of both of these and plug in our bounds. So for a first term, we get the limit as t approaches Negative infinity of negative 1/3 plus 1/3 eat the negative T cubed. Plus for a second term, the limit as t approaches infinity of negative 1/3 e to the negative t over three plus 1/3. And if we look at our first term here and we plug in that limit, we get negative 1/3 plus 1/3 times e to infinity because our negatives cancel out. And if we look at a really quick sketch of the graph of each be X as e to the X approaches infinity. It goes to infinity so this whole term becomes infinity. And since we found that at least one part of our integral goes to infinity and is discontinuous, the entire integral is divergent.

Wen Zheng · Answer

the problem is determined Why that this integral is component or dem urgent. First, we can use use of institution to realise dysfunction as a function which is either to integrate. We can write you is equal to negative one over acts. And then do you and Nico to one over X Squire Jax? So this part is just to be you. Then this integral is the culture e Teo, you hands you and what tax goes to one of you goes to next to one. The axe goes to infinity. One arise goes to zero. This is zero. This is not on improper. Integral. Come. Fine is it seemed integral directly. This&#x27;s equal to you, Teo. Youth power from Nick to one two zero. This&#x27;s he called Tio You two zero one minus You two make you wise power. So this integral is converted. Ondas value is one minus. Ito, make it your wife&#x27;s power.

Fuzail Shakir · Answer

Okay, So after we apply the definition of the improper integral, we end up with 1/2 of the limit as he approaches infinity of the integral from zero to be of X squared plus two to the power of negative two times two x. T s. So applying the power rule for this, we end up with 1/2 uh, times. Oh, yes. This is equal to 15 times the limit as be approaches infinity off one over X squared. Plus two on. We&#x27;re gonna evaluate this from zero. To be applying a fundamental theory of Catholics. You end up with negative 1/2 times the limit as he approaches infinity of one over B squared plus two minus 1/2 minus 1/2. So exploring the limits, you and that 1/2 times won over infinity minus 1/2. So this then goes to Cyril Cyril. Final answer is 1/4

Matthew Allcock · Answer

in this video, we&#x27;re gonna go through the answer to question number 17 from Chapter 58 So asked to evaluate the integral between one and affinity of the natural log of X over X, the X If it exists, it converges. So we&#x27;re gonna do this by parts something so firstly, leaving the not from log of X as it is we then most quiet by the in school of one of X, which is also natural Lagerback&#x27;s. Then we can subtract from the audience girl between one infinity off the derivative off the natural one of X, which is one of rex times by the integral of one of X, which is not a lot of ex, but okay, so now if we call this integral this whole thing that we&#x27;re trying to calculate, I then this is equal to They&#x27;re not your long Max Square and evaluated, but one infinity minus I itself. Because this guy here it&#x27;s just the same as I It&#x27;s so so we can be arranged this to get eye is equal to half size by much. Looking back squared between one and Tennessee, which the huh times the limit be goes to infinity. Uh, the natural log of B squared minus the natural log one squared. Now this much longer be as because infinity diverges therefore, the whole integral diverges.

Amanda Stone · Answer

in this problem were given the improper integral Ellen of acts over x dx from one twin Fendi and were asked to find whether this is convergent or divergent. So I&#x27;m going to rewrite this as the limit as t approaches infinity of 12 tea of Ellen of acts over Axe TX. Okay, so the first thing I want to do is just worry about the indefinite, integral, and then we can plug in our bounds later. So to solve this, we can use U substitution where you equals Ln of acts and d U equals one over acts dx. And that leaves us the integral of you. Do you? And we can right this as 1/2 you squared plus c. And if we go ahead and substitute, are you back in? You get 1/2 hell and x squared plus c. And finally, if we put back in her bounds, we get 1/2 l n of x squared from one to teach you. So from here Ah, we can plug in our bounds and we&#x27;re left with 1/2 Ln of tea squared minus 1/2 l and of one square. This goes zero because Ellen of one equals zero. And if we take the limit of this, we get the limits. As T approaches infinity of 1/2 l and T squared. And if we look at a rough graph of R L and value, we can see that as L and approaches infinity, it goes to infinity, which means that this term is infinity, which makes our whole answer infinity and this means that are integral is dying for urgent

Problem

Determine whether each integral is convergent or …

Problem 13 Medium Difficulty

Determine whether each integral is convergent or divergent. Evaluate those that are convergent.
$$\int_{-\infty}^{\infty} x e^{-x^{2}} d x$$

Answer

The integral converges to 0

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Essential Calculus Early Transcendentals

Anna Marie V.

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Integration Review - Intro

Definite vs Indefinite

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The integral converges to 0

Essential Calculus Early Transcendentals

Anna Marie V.

Heather Z.

Kayleah T.

Kristen K.

Integration Review - Intro

Definite vs Indefinite

James StewartEssential Calculus Early Transcendentals

Anna Marie V.

Heather Z.

Kayleah T.

Kristen K.

James Stewart

Essential Calculus Early Transcendentals