evaluate-the-definite-integral-int_0a-x-sqrta2-x2-d-x-2

Question

Evaluate the definite integral.
$\int_{0}^{a} x \sqrt{a^{2}-x^{2}} d x$

Fuzail Shakir · Accepted Answer

So if we let you is equal to a squared minus x square, we can let negative 1/2 to you is equal to x t X. Then we can rewrite are integral as negative 1/2 times the integral from a square to zero, um, of you to the power over 1/2 you this&#x27;d is just equal to negative 1/3 time&#x27;s you to the power of three over two. Evaluated from a square, eight square to zero. Applying the second part of the fundamental rule of calculus, we end up that this is just equal to a cubed divided by three.

Amrita Bhasin · Answer

we know that Are you? Convene what&#x27;s under the square root, which is a scored minus X squared which means r D X is now negative 1/2 do you over acts which means our bounds are now going to be from a squared to zero because we know a is a squared minus a score which is zero and be a zero squared plus a scored, which is simply just gonna give us a squared. Let&#x27;s let us use the power rule, which means we increase the expert by one and we divide by the new exponents. Now we can simply plug in our bounds. This gives us 1/3 a cute

J Hardin · Answer

Let&#x27;s evaluate the integral from zero a of X squared times the square root of a squared minus X wears now inside the radical we have a squared minus X square. So our tricks up should be X equals a sign data from which we have DX equals Any co sign Taito D data. Since this is a definite or rule, we should see if we could change the limits of integration in terms of the new variable data. So using our truth substitution, plugging and X equals zero. Yeah, zero equals scientific data. Excuse me? A signed data. And also, when we do it in a girl, a trick some of this form. It will always put the condition that data&#x27;s betweennegative over too, and powerful too. So this is for the a sign data tricks up. So here we have a signed data equals zero. So assuming that is down zero, we have signed equal zero, and the only time that happens in this honorable zero. So that&#x27;s our new lower limit for the upper limit. We have a so plugging in a for X with a equals a signed era, so divide both sides by a Scient ADA equals one, and the only time that happens in our interval is that over too. So those are new limits of integration and zero and power too. So before we agree, right this original problem, let&#x27;s just simplify this term right here. Square root of a squared minus X squared. So that&#x27;s a square science. Where data here I could pull out of a squared or left over with one minus sign square. So we had eight times Square root co sign squared data equals Saeko San Diego. So plugging all this in we could write Original problem is the inaugural from zero to pie over too. X squared becomes a squared sine data sine squared data and then the radical. We just simplified this. This&#x27;s a co sign data and then DX was also a co santa. So here, let&#x27;s come. Let&#x27;s simplify this So we have AIDS to the fourth power. Let&#x27;s pull that constant outside the integral. And then we have science where times co sign squared. Now this is a trick in a rule, and since we have even powers on both side and co sign will want to use the Haft angle identities, science square is one minus coastline to data over to and co sign squared one plus coastline to date over, too. So let me go to the next page, since I&#x27;m running out of room so multiplying those two&#x27;s together we get a four on the bottom and then one minus co sign to data one plus goes into We could re simply by this angel. Fourth over four one minus co sign squared potato And then using the Pythagorean identity, we can write this assign squared two There now, once again using the half angle formula Weaken right sine squared to data is one minus co sign of two times to Daito. So it&#x27;s a fourth Ada over soon, and I should be consistent. This&#x27;s a definite liberal and from the previous page our end points for See you in a pile over too. So we still need those now replacing sine squared with this one minus co sign Fourth eight over, too. Went into the fourth over eight in a girl C on a paper too one minus coast. And for data And this we know how to number eight. This is eight of the fourth over eight in a roll of one is data in a World Co sign for theta designed for theta over four. So now we plug in the end, points into this expression went into the fourth over eight pie or two minus sign of four times Pirates Who it&#x27;s to pie. And then when we plug in zero, we have zero minus sign zero over to over four. So these terms are both zero sign of two pi zero. So we&#x27;re left over with a the fourth over eight times. Pyro, too, which we can write his pie into the fourth over sixteen, and there&#x27;s our final answer.

Amy Jiang · Answer

Okay, Senator. Sent in are integral. We have X squared and X so we can set you to be equal to X squared plus one. And then we have Do you that&#x27;s people to actually shouldn&#x27;t be a power of two. Here, Um, he used you. Put two x, the X. And now let&#x27;s divide by two on both sides so we can get xcx, which is part of our integral. Okay, so this gives us the integral of X dx is the u over to and then we have the square root of X squared plus one which is equal to you and Allah to replace our endpoints up as well. You know that, Um, when accident with the to you is equal to two squared plus one, so that&#x27;s equal to five. And when X equals zero, you have you is equal to zero plus one. Judy, go to what? So we have from 1 to 5. Okay, so that gives us you through the power of 1/2 plus one. So that&#x27;s 3/2. And in two of the three here evaluated at, um actually, we still have this, um, 1/2 aka. So this two turns into it one and then we evaluate this up five and one. So this gives us 1/3 you, which is five to the part of the over two and then minus one to the power of 3/2. And it&#x27;s also include 21 so we get the falling for a solution.

Amrita Bhasin · Answer

okay, We know it could consider Are you to be what&#x27;s under the square root, which is X squared plus a square, which means our d&#x27;you is to axe d x, which means x d x is 1/2. Do you only transfer the two over to the other side? Which means we can now pull out the constant of 1/2 and then we have are lower bound is zero squared plus a scored, which is a scored in the upper ground is a squared plus base chord which is two squared and we have the squirt of you. Do you? Okay? Take the powerful. Which means increase the exponents by one and divide by the new exponents. Now, using the fundamental damn of calculus, we know we can plug in.

Amy Jiang · Answer

Okay, So it&#x27;s not for noting noticing that we haven&#x27;t x squared here, and we know that the derivative of that is in terms of X. So we consent you to be with your X squared plus one, and I&#x27;ll do you that&#x27;s able to two x the X on. Now let&#x27;s move this too. So we have X dx in our original integral. So we get in a girl of xcx. Let&#x27;s see you off to and then one over the square root of you. Okay? And I was also, um, find our endpoints on X is equal to one. You was equal to one plus one, which is to when exit you could throw. We have you. Is it? Which one? That&#x27;s from one thing to Okay. And I&#x27;ll be right. Um, this integral as negative exponents. So that&#x27;s you. Too far. Negative 1/2. Do you are not taking our inner girl. We get 1/2 you could&#x27;ve are negative 1/2 plus one. So that&#x27;s 1/2 over 1/2. So I get you one or two violated a two on one that is we have you to power or two to the part of 1/2 minus. Wanted about 1/2. It&#x27;s not people to the square root of two minus one

Problem

Evaluate the definite integral. $\int_{0}^{1} \f…

Problem 48 Medium Difficulty

Evaluate the definite integral.
$\int_{0}^{a} x \sqrt{a^{2}-x^{2}} d x$

Answer

$\frac{a^{3}}{3}$

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Integration Review - Intro

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$\frac{a^{3}}{3}$

Biocalculus Calculus for the Life Sciences

Catherine R.

Kristen K.

Samuel H.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

James StewartBiocalculus Calculus for the Life Sciences

Catherine R.

Kristen K.

Samuel H.

Joseph L.

James Stewart

Biocalculus Calculus for the Life Sciences