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

$$7 / 9$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Missouri State University

Harvey Mudd College

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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Evaluate the definite inte…

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Evaluate the given definit…

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Evaluate the integral.…

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So as you look at this problem or doing the integral from 0 to 1 of x times the quantity you know, I'm glad they didn't take the square root, but they could have, uh, this three x squared plus one, uh, to the one half powers. The same thing as the square root. Uh, d X. So what I would do is utilize u substitution. We're looking at three x squared plus one. The quantity, it's inside the parentheses. So then d you would equal six x d x And what I like to do is sell for DX by multiplying by +16 Um, 1/6 X, I should say over here. So now I'm ready to, uh, do this anti derivative. Well, let's rewrite it in terms of you. So in terms of you, you have to change your balance. Plug in zero for X three times zero is still zero plus one. So now it's going to be the integral from one and plug in one square. It is still one times three is four plus one. I said there are three times. One is three plus. One is four. I like to leave that X there. There's a reason for it. Rewrite this as you to the one half power and I'm going to replace DX with that 1/6 x d you? The reason why I like that is I don't cancel out these exes now. Everything is in terms of you and it makes perfect sense to me. I'm ready to do the anti derivative, which is adding 1 to 1 half that's three halves multiply by the reciprocal of three house, which is two thirds. Don't forget about that 16 either. So we have this from 1 to 4. Now, you can also factor out that that piece so you can just look at that and say Okay, well, to over six would reduce to one third. So we're looking at 1/9 times a quantity. Well, if I plug in my bounds, this means the square root of four cube, while the square to force two cubed is eight to tap suit jumpsuit and then plug in one. Well, one day any power is just one. It should make sense because you're doing the square. That one is one cubed 12 Excellent. Everyone is still one and then subtract off those values. And, uh, we're actually ready to finish this problem because you just have eight minus one at seven and put that into the numerator seven times. One is seven overnight.

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