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Evaluate the given integral.$$\int_{0}^{2} \frac{1}{\sqrt{4 x+1}} d x$$

$$1$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 7

Substitution and Properties of Definite Integrals

Integrals

Campbell 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.

02:27

Evaluate the definite inte…

03:35

Evaluate the integral.…

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so forgiving the integral from 0 to 2 of some function one over the square root of for explosive one. What we want to do is use U substitution and let that equal the four X plus one. Because that's your group and simple So then you can look at the derivative of that which is just before D s. And in my experience, my students need to see that D X is equal to 1/4. Do you? That way, when we rewrite this integral, we can look at that and say, Okay, well, how about instead of writing one over the square thio you because you is equal to four X plus one, I can rewrite that as to the negative one half power, so use replacing this piece. But if it's in the denominators of negative exponents in the square, root is the one half hour and I'm gonna replace DX with 1/4 d you, I'm just gonna move. I don't like how I wrote that. I'm gonna move the 1/4 in front, and I need to change my bounds as well, because this is when x zero, when you're changing things, Thio you so four times zero plus one gives me a you value of one and plugging in to four times two is eight plus one is nine. So now we can do the anti derivative of that, Uh, where you add one to your ex con it. Mega. One half plus one is positive. One half multiply by the reciprocal eso are multiplying and 1/4 by two. Well, that gives me one half from 1 to 9. So now I can plug in those bounds. Well, these remember this means the square root of nine, which is three. And then this is the square root of one which is one and that half is still there. That's why I still just wrote three has minus one half, which equals two halves, which equals an area of one.

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