determine-whether-each-integral-is-convergent-or-divergent-evaluate-those-that-are-convergent-int_0i

Question

Determine whether each integral is convergent or divergent. Evaluate those that are convergent.
$$\int_{0}^{\infty} \frac{1}{\sqrt[4]{1+x}} d x$$

Amanda Stone · Accepted Answer

in this problem were given Thean proper integral one over the quantity one plus x two, the 1/4 from zero to infinity and were asked to determine whether this is a convergent or divergent Integral. So to do this, I&#x27;m going to take that infinity bound. And instead I&#x27;m going to write it as the limit as T approaches infinity of t 20 of 1/1 plus x two, the 1/4 DX and now we can solve this. So I&#x27;m gonna go ahead and begin by just worrying about our indefinite, integral installing for that and then we can plug in the bounds. So if we solve this, we get 4/3 times the quantity one plus x to the 3/4 plus C. If we go ahead and plug in our definite bounds into this, we get 4/3 times a quantity one plus x to the 3/4 from zero to t. If you go ahead and plug in these bounds, I&#x27;m going to factor out that 4/3 and put it in the front and we get the quantity one plus t 3/4 minus one. So now all we have to do is take the limit of this as T approaches infinity of 4/3 times one plus t 3/4 minus one And if we do this will notice that we can also right this as 4/3 times infinity to the 3/4 minus one. And this is just going to leave us with infinity. And because of that, we know that this integral has to be divergent.

Wen Zheng · Answer

the problem is determine whether this into grew is compartment or there wouldn&#x27;t. This is improper. Integral. The high definition. This is ICO. Choose limit. He goes to infinity on DH integral of one over one last axe. You unforced power from their own? Sure, Jax. So we can realize dysfunction as cost you girl, They&#x27;re on TV. Suit one plus x Teo make you one force power. Yeah, Thanks Then we need to find anti derogative off dysfunction. This is a iko too, Lim. It goes into infinity indeed. Grow aunt had already with this function is one over minus one over or us one one plus ice. You next to one force. How long from year old, then? Climb T and they&#x27;re all here. This is ICO, that limit He goes to infinity. This one over three or four. But this is this is war over three. And this is one last read. All four minus with three Cam&#x27;s one. This is the power ofthe one plus two is positive. Once he goes to infinity one plastic to three ofthe force Palmer goes to infinity with other is infinity thiss into grow as the word and

Fuzail Shakir · Answer

Okay, so we can rewrite this as negative two times the limit as be approaches infinity of the integral from one to be want to be of e to the power off Negative the square root of X times one over. Let&#x27;s see one over Negative too. Times the square root of X t X. All right, so if we apply the integration girls, here we end up with this is eagle to negative two times the limit as he approaches infinity off the to the power of negative squared of X. And we&#x27;re gonna do this from one to be applying a fundamental theory of calculus. This is equal to negative two times the limit as be approaches infinity of e to the power of native square to be minus each of the power off. Negative one evaluating the limit. This is equal to native To have each of the power of negative cleavage is one over each of the power infinity minus eat the power. A support to this therefore goes to zero. So our final answer is to overeat

Wen Zheng · Answer

The problem is you tell me why this integral is convergent where there weren&#x27;t. If you don&#x27;t know howto find its anti derivative of this function, you can think about my thereto u substitution. But this question we&#x27;re kind like you is equal to U until Max Then is he going to you? Squire? And the axe is it got you to you. You Now this nd grow is equal to you You, you over you us, you squire. I will see you. And from well, I&#x27;m Tio infinity. We can kinds out you So this is called Chu integral from one to infinity. Two times you over. One pass Useless now by the definition of an improper into your o This is the code to the limit he goes to in planete into girl from one to t The function is who the over one plus you square And this is your coat too, Lim. He goes to vanity and anti derivative of the function One over one plus your square is our tenant. You This is a culture too. Terms, lieutenant, you from one two You this you come too. Two hams. Our tenant You minus ten in one. That&#x27;s what he goes to infinity. I&#x27;ve tended to you goes too high over to answer if you come too. Two towns. I already too minus octane in one. If we go too Hi over, for his answer is you go to hi Over two this integral has converted and the value is high over too.

Amanda Stone · Answer

in this problem were given the improper integral Ellen affects over the square root of X T X from 0 to 1 and were asked to determine whether this integral is convergent or divergent. So in this problem, if we plug in zero into this integral, it&#x27;s discontinuous. So I&#x27;m going to rewrite this as the limit as T approaches zero of t tow one of our function. So I&#x27;m gonna just worry about the indefinite integral first and solve that and then we&#x27;ll pulled in our bounds. So if we do that will notice that will have to use integration, buy parts to solve this. So we&#x27;re gonna have to find are you and our Devi and then we can find RV. And do you so a little trick that I was taught in identifying what is your you and what is your D. V? Is this acronym L I A t e. Where else stands for log functions. I inverse drink functions a algebraic functions, t trig functions and e exponential functions. So the trick is going from left to right of this acronym which ever type of these functions appears first in your integral. That&#x27;s your you and then everything else including your D X is your Devi. So in this case, we do have a log which means that are you Value is going to be Ln fx And that means everything else is going to be our devious or devi is going to be the square root of X d. X. So we go had installed this Ah, the rest of the things in our little chart here will get do you to be one over x dx and V to be two x to the 1/2. So now always to do is use the formula for integration by parts which is UV minus the integral of V d U. And that will give us two x to 1/2 Ellen of X minus the integral of to the X to the negative 1/2 or two times x to the negative 1/2 d x If we go ahead, solve this integral. We get two x the 1/2 Ellen of X minus four X to the 1/2 and then if we go ahead and add in our bounds, we evaluate this from t 21 I&#x27;m going to go ahead and plug in our bounds and take the limit of this. So the limit as T approaches zero of to Ln of one minus four minus two times. Teach the 1/2 Ellen of tea minus four t to the 1/2. So right away, we noticed that if we plug in zero this term become zero and then Ellen of one is zero said this become zero. So all we have left to worry about is negative for minus the limit as T approaches zero of two. T two, the 1/2 Ellen of tea. So here we have a little problem, Ellen of tea. If we pulled in zero Ellen of zero is undefined. It&#x27;s indeterminate. So in order to solve this term, the to teach the 1/2 Ellen of tea we&#x27;re going to use lay off those rule so we can take the limit as t approaches zero of Ln of tea over to t to the negative 1/2. That&#x27;s just another way of writing teach of 1/2. We can write it as one over teach the native 1/2 and if we take the derivative of the top and bottom, we get the limit as T approach zero of one over tea over t to the negative three halves. And if we simplify this, we get the limit as T approaches zero of t to the three halves over T if we just take the reciprocal of both of those and then we can rewrite this as teach the 1/2 which if we put in zero, that goes to zero. So now we know. But this term goes to zero. And that leaves us with our answer being negative four, which means that this integral converges to negative four.

Amanda Stone · Answer

in this problem were given the improper in a girl one over X squared plus X dx from one to infinity and were asked to find whether this is convergent or divergent. So I&#x27;m gonna rewrite this as the limit as t approaches infinity from one to t of our function. So I&#x27;m going to begin by just focusing on our indefinite integral one over X squared plus X dx, and then we can add back in our bounds. So in this problem to solve this in a row, we&#x27;re gonna have to use partial fraction decomposition. So that means that we can look at this function as one over X Times X plus one and so we can rewrite one over X squared plus X as a over X plus B over X plus one. If we multiply both sides by X squared plus x, that gives us one equals a times X plus one plus be X, and we can again rewrite this as one equals a X plus a plus be X. And then we can group are excess together and write one equals a plus B X plus a. So in this equation, uh, a equals one because there no X values on the left side, and then we can write zero equals a plus B and using what we already knows A equals one, we can prevent it and get zero equals one plus B and B equals negative one. So we unplug these values of A and B back into our equation for partial fraction decomposition. And we get one over X squared plus X equals one over X minus one over X plus one. And now we can take the integral of both of these. So if we do that, we get l N of X minus l and of X plus one, all evaluated from one to t. If we plug in our bounds, we get Ln of t minus l and off one minus. Oh, on of t plus one minus. L End of Chu. We know that Ellen of one goes to zero so we can just cancel out that term. And now I&#x27;m going to take the limit as t approaches infinity of l end of tea minus l end of t plus one plus Ln of to And if we plug in infinity to this we get infinity minus infinity plus L one of two. And we know this because if we look at a quick graph of Ellen of X as Ellen of X and purchase infinity, it goes to infinity because we have infinity minus infinity that cancels out and we&#x27;re left with Ellen of To which means that are integral is convergent.

Problem

Determine whether each integral is convergent or …

Problem 6 Easy Difficulty

Determine whether each integral is convergent or divergent. Evaluate those that are convergent.
$$\int_{0}^{\infty} \frac{1}{\sqrt[4]{1+x}} d x$$

Answer

integral is divergent

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James Stewart

Essential Calculus Early Transcendentals

Anna Marie V.

Caleb E.

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Integration Review - Intro

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integral is divergent

Essential Calculus Early Transcendentals

Anna Marie V.

Caleb E.

Kristen K.

Michael J.

Integration Review - Intro

Definite vs Indefinite

James StewartEssential Calculus Early Transcendentals

Anna Marie V.

Caleb E.

Kristen K.

Michael J.

James Stewart

Essential Calculus Early Transcendentals