if-a-and-b-are-positive-numbers-show-that-int_01-xa1-xb-d-xint_01-xb1-xa-d-x

Name: if a and b are positive numbers show that int_01 xa1 xb d xint_01 xb1 xa d x
Uploaded: 2020-03-29
Duration: 47 s
Channel: Numerade
Description: If $a$ and $b$ are positive numbers, show that $$\int_{0}^{1} x^{a}(1-x)^{b} d x=\int_{0}^{1} x^{b}(1-x)^{a} d x$$

Question

If $a$ and $b$ are positive numbers, show that $$\int_{0}^{1} x^{a}(1-x)^{b} d x=\int_{0}^{1} x^{b}(1-x)^{a} d x$$

Amrita Bhasin · Accepted Answer

this question asked us to proved the given statement given the A and B are positive numbers. So in other words, from our bounds from zero toe one, we can flip these to make our bounds from 1 to 0. Given this, we know we can, right? Negative 120 you to the B terms one minus you to the A m nd you as equivalent to from 01 you to the B times one minus you too, eh, D&#x27;You Therefore, given the fact that you could be replaced with axe, we know we can write this as from 0 to 1 X to be so we&#x27;re replacing every you with acts one minus x to the power of A and R D U becomes d axe. This is proven the statement.

Dakarai Holcomb · Answer

So we&#x27;re going to show that this following equation is true. When A and B are positive numbers in order to do that first we&#x27;re going to let a equal one and be equal to two. We&#x27;re going to solve for the first part of our equation. So we&#x27;re going to do are integral From 0 to 1 of x rays, that one times one minus X. All this, raised it to the X. So next year you&#x27;re going to want to factor this and break this down our middle portion so that we end up with x minus X squared sorry not X squared two X squared plus X cubed dx. So when we take the integral of each one of these individual parts, we&#x27;re going to end up with one half minus two cubed plus 1/4 Which equals 1/12. All right. So now that we&#x27;ve shown that the second half or sorry the first half of the equation is equal to one half. We&#x27;re going to do the same thing for the second half of the equation. Mhm. So we have the integral of 0-1 X squared Parentheses, -X. Um dx So then we&#x27;re going to rewrite this integral of 01 as X squared minus execute and we get this from distributing the X square to our parentheses the X. Now when we take the integral of each one of our components Here we end up with 1/3 minus one force. This is after you take the integral of each component and plug in your one. This will also give us 1/12 showing that our equation is indeed true when our A and B is possible.

Pravakar Paul · Answer

we&#x27;re going to use substitution. To solve this problem, we have to show that integration from 0 to 1. It&#x27;s cute. Then one minus extra the poor. It t X stamos integration from 0 to 1 extra deport eight times one minus X two Cube DX safely. Just look at the problem closely. The X and one minor six has been interchanged in the second expression. So let&#x27;s start with the first expression for the integration from Syria to one excuse that one minus extra. DeBoer it ticks and we&#x27;re going to substitute a new variable. So let&#x27;s take Why&#x27;s he called to one minus six? If I take why&#x27;s it called one minor sex, then tee off. Why is the culture my negative DX? So this implies D X is equal to my negative, Do you like? Okay, so let&#x27;s try to calculate the new limits. Now the newly tells us so when excess sickle to zero. This implies y sickle to one on when excess tickle toe won this in place. Why&#x27;s it called zero? Okay, so let&#x27;s go back and try to solve this problem. So the new limits will be no. 1 to 0 because the limit has been interchanged on. What is my ex exes? Exactly? One minus y. So let me just write that down too. So why is it called Tau One minor six? This implies X is equal to one minus, like so. This is kind of symmetric. So X is one minus y, then Cube on one minor success. Why? So this becomes why? To deport eight. And what is my DX dy access Negative D y. So I can pull that negative sign outside. So this becomes negative integration from 1 to 0 one minus y cube. Why? To deport it. Do what? So now if I intercede and stop on limited lower limit on the lower limit all of our limit it introduces a new negative sign. So this becomes negative off negative integration from 0 to 1 now one minus y cube. Why? To deport eight. Doing on, then to negative sign exit. A positive sign. So it tells me this is saying that 0 to 1. Why two before it followed by one minus y cube delight. Now, in integration, we can always change the variable. So this becomes 0 to 1 extra before eight times one minus X Cube T X and that&#x27;s exactly that. Lift and sight. Sorry. Right hand side up the integration. So this is proofed. Okay, so let&#x27;s move on. Let&#x27;s go to the second problem to second Problem says I have this equality 0 to 1. Next to the bar, one minus extra. The Barbie D X is a call to integration from 0 to 1 X to the Barbie. Then one minor, six to the board ticks. Okay, Second, we&#x27;re just going to use the same transformation. So why is going to be my one minor six? This implies my ex is goingto one minus y on T X is a call to negative. Do I on dhe When X is equal to zero, my wife is equal to one. When X is equal to one, my wife is equal to zero. So that right in sight becomes integration from 0 to 1 next to the body one minus extra. The Barbie T X, which is same as integration. Now this becomes integration from 1 to 0 extra deportee. This is simply one minus y to the party on one minor Sixtus. Why? So this is just white or the part B and he excess negative, Do I? Okay, so this becomes negative integration from 0 to 1 one minus y to the party, followed by way to the party, then do I? And again excuse me, I changed. Ain&#x27;t limit, Phyllis. Write it properly. So this becomes integration 1 to 0. And finally, if I teens the limit off the integration, then it introduces a extra negative sign. So this becomes negative off negative integration from Syria to one. Then why? To the part B one minus y to the party Dubai and two negative exit positive. So finally we have 0 to 1. Why? To the Barbie, then one minus y to the power He&#x27;s d y on in the definite do grill, we can always change the new variable eso it becomes 0 to 1 x to the power P times one minus x to the party on times DX, which is precisely the lift in sight off that integration or two hypothesis on

R M · Answer

problem. 74 Integration off export and sine inverse X The X, which is equal to U, is equal to sign Angers X the you is ableto B X over one minus X squared. Needy is equal to export and the X the is able to export and plus one over and this morning. So this is equal to sign Angers X and yes, times export and plus one over and plus one minus generation of X. And that&#x27;s one over and plus one times he x over square root off Omanis X squared, which is equal to export any plus one times. Signed English X over and plus one minus one over M plus one figuration off X n plus one over square root off one minus X squared years.

Yaw Asomani · Answer

gonna prove that this is bad. This equation is about it, and it is just simple. We&#x27;re gonna use the, uh, definition for definite intervals. So when you have a b angsty eggs, we just gonna evaluate the integral What does a right, uh, the integral off x d x is just x squared over two, right? And then you just have upper limit and lower them in here and in definition of different intervals is just evaluating. A definite answer is just upper limit minus tournament right by the fundamental to theorem of calculus FTC this half would be minus half a All right, so we&#x27;re just using the FTC two fundamental theorem of calculus where this f is the integral off is the indefinite integral off this little F and this little half in our cases, uh, eggs, right? And so this indefinite integral of that is just X squared over two. So we need It&#x27;s just f B minus half a favor. Just gonna put a B and es. So effort be is gonna be B squared origin minus ever a. It&#x27;s just a square over, too. And then you can combine it into one B squared, minus a square, and that completes the proof

Michael Jacobsen · Answer

in this video, we&#x27;re going to be calculating the indefinite integral of x times beat of the X DX. First notice that the base be is unspecified, that we still have certain requirements for being a the X To be defined be has to be positive And in order for this problem to be interesting be should not be equal toe one since one of the power of X is just one now, to get started on this problem, we can solve it by using the method of integration by parts. Let&#x27;s set you to be equal to X so that do you must then be DX next. All that&#x27;s left for Devi is B to the power of X times DX So we need the anti derivative of BX. Well, we know that that derivative with respect to X of B to the power of X is itself times the natural log of be. So if we&#x27;re taking the anti derivative we Formby to the X divided by the natural log of B, Now we&#x27;re ready to begin that integration by parts solution first, take u times V so that this will be equal to all together. X times B to the power of X divide by the natural log of B, then subtract a particular anti derivative formed by taking V times to you. In this expression, one over the natural log of B is concerned constant Sophie like we can place it here just to make it less of a distraction in the next indefinite integral, which will be Peter the X DX. On the next step, we see that this is equal to first copy X Times view the X over the natural log of B minus. Now the anti derivative of B to the power of X. We&#x27;ve in fact, already done once. It&#x27;s this expression here, since we already have natural rugby and a denominator will be race of power to so we have B to the X divide by the natural log of B squared. Then we throw in an arbitrary constant to get the most general anti derivative. And if we prefer, this can be cleaned up a bit by, say, factoring out B to the power vex. That would leave behind X divide by the natural log of B minus one. Divide by the natural log of B squared, plus a constant

Problem 67 Hard Difficulty

If $a$ and $b$ are positive numbers, show that $$\int_{0}^{1} x^{a}(1-x)^{b} d x=\int_{0}^{1} x^{b}(1-x)^{a} d x$$

Answer

$\int_{a}^{b} x^{a}(1-x)^{b} d x=\int_{a}^{b} x^{b}(1-x)^{a} d x$ Proved

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James Stewart

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Grace H.

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Integration Review - Intro

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$\int_{a}^{b} x^{a}(1-x)^{b} d x=\int_{a}^{b} x^{b}(1-x)^{a} d x$ Proved

Biocalculus Calculus for the Life Sciences

Grace H.

Catherine R.

Kristen K.

Samuel H.

Integration Review - Intro

Definite vs Indefinite

James StewartBiocalculus Calculus for the Life Sciences

Grace H.

Catherine R.

Kristen K.

Samuel H.

James Stewart

Biocalculus Calculus for the Life Sciences