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If $ a $ and $ b $ are positive numbers, show that$$ \int^1_0 x^a(1 - x)^b \,dx = \int^1_0 x^b(1 - x)^a \,dx $$

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01:51

Frank Lin

01:05

Amrita Bhasin

02:39

Linda Hand

Calculus 1 / AB

Chapter 5

Integrals

Section 5

The Substitution Rule

Integration

Khalid A.

November 16, 2021

chapter 11

sequences

Justin I.

January 31, 2021

If f is continuous on [0, ????], use the substitution u = ???? ? x to show that

Missouri State University

Oregon State University

Baylor University

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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So we're going to show that this following equation is true. When A and B are positive numbers in order to do that first we're going to let a equal one and be equal to two. We're going to solve for the first part of our equation. So we'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'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're going to end up with one half minus two cubed plus 1/4 Which equals 1/12. All right. So now that we've shown that the second half or sorry the first half of the equation is equal to one half. We'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'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.

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