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Evaluate the given integral.$$\int_{0}^{1} \sqrt{5 x+4} d x$$

$$38 / 15$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 7

Substitution and Properties of Definite Integrals

Integrals

Harvey Mudd College

Baylor University

University of Nottingham

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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Alright, we're gonna do you substitution for this problem. The integral from 0 to 1 of the square root of five X plus four DX. So I tell my students to look for a grouping symbol on the square. Root is a group, so I'm gonna do everything that's under the square root and then do you the derivative of that would be five g X and in my experience, my students need to rewrite this as dividing the five over. The reason why that's helpful, I think, is because then students can say Okay, well, this is equal to the integral. I'll mess with the balance here shortly. I could replace five X plus four with you rewrite to the one half power and then I could replace DX with 1/5. Do you on dso again? My bounds need to change. So when I plug in zero for X factor 00 plus for US four and then when I plug in 15 times, one is five plus four is nine and now I'm ready to do the anti derivative, which is adding one to that exponents that's three halves multiply by the reciprocal of three halves um, which would be two thirds. But don't forget about this five. So you have to do three times five is 15, and that's from 4 to 9. So now you plug in your upper bound Well, this means the square root cube. So the square to nine is three. Cubed is 27. We're looking at 27 times to his 54 15th and then minus again, plugging in forward to square to force to Cuba's eight times. This, too, would be 16 15th. And now that the denominator with same we can just put them together and 50 for minus 16 is 38 that is your final answer.

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