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Evaluate the given integral.$$\int_{-1}^{0} x^{2}\left(2 x^{3}+1\right)^{5} d x$$

$$0$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 7

Substitution and Properties of Definite Integrals

Integrals

Campbell University

University of Michigan - Ann Arbor

Boston College

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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02:54

So we're going to use u Substitution do this problem. We're talking about the end of roll from negative 1 to 0 of X squared times that quantity okay to execute plus one to the fifth power DX. Now I always tell my students when you're doing you substitution toe, look for parentheses. I'm looking for to execute plus one That's the parentheses. So when I do, do you bring the three in front? We're taking the derivative. So two times 36 x squared dx and what I typically do is have my soon sulfur DX, which would be 1/6 x squared. Do you eso as I go back to the derivative? Uh, first of all, I leave X squared alone and I leave the to the fifth power alone. What I do is I replace that to execute, plus one with you. And then I replaced DX with what that equals two. Which is the 1/6 x squared. Do you? And then my students actually see the X squares go away on then. From there, I also have to change my bounds. Wolf, I plug in negative one to the third. Power is negative one times two is negative. Two plus one would be negative one. I'm plugging in zero not to do my upper bound her ex eso when x zero, Why is one And now this is kind of an easy problem because now what we have is just an odd function. This is an odd function. Yeah, I could spell And the bounds are equal distance from the origin. Eso, without doing any math already know that the area below the curve in the area above the curve will cancel eso. Therefore, I already know this answer will be zero. But if you don't believe me, you can go through the work where you add one to your exponents. Divide by your new exponents. Will, six times six is 36 um, plus C. But no, no. Plus, you hear from negative 1 to 1 because it's a definite interval. So if you plug in one into that problem, uh, and then if you plug in negative one, you get another one and 1/36 minus one for 36 equals zero. Exactly what I said the answer would be

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