determine-whether-each-integral-is-convergent-or-divergent-evaluate-those-that-are-convergent-int-15

Question

Determine whether each integral is convergent or divergent. Evaluate those that are convergent.
$$\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}$$

Amanda Stone · Accepted Answer

in this problem were given the improper integral de axe over the square root of one minus X squared from 0 to 1 and were asked to determine whether this in a role is convergent or divergent. So since this integral is discontinuous, X equals one. We can rewrite this as the limit as t approaches one from zero to t of our integral. So I&#x27;m going to begin by just solving the indefinite integral D X over one minus x squared. And this is actually the formula for sine inverse. So this is just sign in verse of X plus e, and we can now rewrite this with our bounds. That would be sine inverse of x from zero to t. If we plug these bounds in, we get sine inverse of t minus sine inverse of zero. And finally, we can take the limit of this the limit as T approaches one of sine inverse of tea minus zero, because sine inverse of zero is just zero. And the sine inverse of one is high house, which means that are integral converges to high halves

Wen Zheng · Answer

problem is determined why the h Integral is convergent or divergent evaluated Those that are convergent this improper integral By definition, this is equal to the limit. A goes to one from the homicide as integral is from 0 to 80. Yes. Oh man, you would have one minus x choir. We compute. It&#x27;s definitely integral first, so there&#x27;s definitely integral. This is the echo two oxide X from zero to a. This is equal to oxide a minus sign zero. This is minus zero. We know what a goes to one axon A goes to hi over to so full this integral this integral is converted and the value is high over to

Wen Zheng · Answer

the problem is determine whether this into grow is convergent or that wouldn&#x27;t before with self this integral the first we can use you substitution toe compared this function function that easier twenty grow we can write you. If he called to one plus Act two Reese power then you if Iko too three ham&#x27;s too square the axe you then this anti girl is ICO too So x squired d x This part is equal to one third you This is one over you to half on DH hams one third do you on one acts goes to zero you cause to want what axe goes to Bennett e You also goes to infinity the cost of energy and then use the definition of improper integral this ego to limit he goes tio financially on DH integral one third ham&#x27;s one over are we could right dysfunction as you and Teo negative one half from one to and then find it&#x27;s anti derivative Off this function, this is you cut to the limit, he goes to infinity and this is a constant number one third and aunt derogative of you to make you one half. This is E Coto one over. Negative one. Have Ruslan. How&#x27;s you? You make it one half a swan from one to then planted in key and one here. So this is you. Coach is limited. He goes to infinity one third times. This is the one over one half. This&#x27;s two ham&#x27;s You to one half from one to on three Lim. He goes to infinity, my third to third hams he to one half minus one. The thing&#x27;s monte goes to infinity. He to one half goes to infinity. So the answer is infinity. That&#x27;s indeed, girl, that merchant.

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

Wen Zheng · Answer

the problem is getting wider. Each integral is converted over skyward and wild about the tales that converted base one acts goes to next. One. Dysfunction. The denominator of dysfunction goes to zero. This&#x27;s an improper, integral after mission. This is equal to Lim. A toast to make one from right hand side Andi into grow is a from a too two act&#x27;s over past one squire the ex. So we computed the definite into zero first. But this definitely eating girl we can read. Write. It&#x27;s a function X plus one over us. One squire, minus one over express one square The ice. Yeah. Then this&#x27;s a yuko to Ellen. Ex us Juan from a too minus negative one over X paswan from a tutu, This is Echo Two, Ellen three minus. How in a cross want Andre? Here&#x27;s this in class one third, minus one over a plus one. We can read like his dysfunction eyes How in three US one third, minus X plus one One here is a eh Right here a swan. How plus one A swan over a swan. Look at this function. When a goes to next to one a plus one goes to zero. It&#x27;s a press want times how in a swan. But this dysfunction it goes to zero. So the whole function goes to infinity. So this interview is that word?

Amanda Stone · Answer

in this problem were given the improper integral X squared over the quantity one plus x cute all square root in D X from zero to infinity and were asked to see whether this integral is convergent or divergent. So I&#x27;m going to rewrite this as the limit as T approaches infinity and change my bounds to zero and t of our function. So I&#x27;m going to begin by just looking at this indefinite integral in solving for that. So to solve this, we&#x27;re going to have to use U substitution where U equals one plus x cubed and D u equals three x squared d x. And if we plug, Are you substitution in? We get a 1/3 outfront and we get a one over the square root of you, d you. And if we go ahead and solve this in a girl, we get 2/3 times you to the 1/2 plus c. And if we go ahead and substitute back in one plus x cute for you, that gives us two third&#x27;s times one plus x cute, the 1/2 from zero to t. So from here we can plug in these bounds on going to factor out the 2/3. And that leaves us with 2/3 times the quantity. One plus t cute to the 1/2 minus one. And all we have to do now is take the limit of this. So we get the limit as T approaches infinity of 2/3 times one plus t cute to the 1/2 minus one. And we&#x27;ll notice that this term goes to infinity because this becomes infinity to the 1/2 which is just infinity. And this makes our entire limit infinity, which means that this integral is divergent.

Problem

Determine whether each integral is convergent or …

Problem 30 Medium Difficulty

Determine whether each integral is convergent or divergent. Evaluate those that are convergent.
$$\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}$$

Answer

the integral is convergent to $\frac{\pi}{2}$

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James Stewart

Essential Calculus Early Transcendentals

Catherine R.

Anna Marie V.

Caleb E.

Michael J.

Integration Review - Intro

Definite vs Indefinite

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the integral is convergent to $\frac{\pi}{2}$

Essential Calculus Early Transcendentals

Catherine R.

Anna Marie V.

Caleb E.

Michael J.

Integration Review - Intro

Definite vs Indefinite

James StewartEssential Calculus Early Transcendentals

Catherine R.

Anna Marie V.

Caleb E.

Michael J.

James Stewart

Essential Calculus Early Transcendentals