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Given $\int(2 x+3)^{2} d x,$ (a) evaluate this integral by first multiplying out the integrand, and (b) by substitution. (c) Explain the different results.

(a) $\frac{4}{3} x^{3}+6 x^{2}+9 x+c$(b) $\frac{1}{6}(2 x+3)^{3}+C$(c) They differ by a constant.

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 3

The Substitution Method

Integrals

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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 integral.…

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Evaluate.(a) $\int_{2}…

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Find the following integra…

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Evaluate(a) $\int_{1}…

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Evaluate by Substitution

Evaluate Integral

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Evaluate the integral usin…

we have the integral of two x plus three squared, and there's actually two ways to solve this problem. The first way, which is probably the easiest, is to apply a use substitution by taking you to be two X plus three. That way do you over to is equal to DX, and we can rewrite this integral as one half times the integral of you squared, which is equal to 1/6 times you cubed, which is, um, which now we replace you by two x plus three and then add C. So that's the first way. The second way to solve this problem would be to multiply this expression out, which is straightforward. If we do that, we get the integral of four X squared plus 12 X Plus nine, which is equal to four thirds X cubed plus six X squared plus nine X plus C. And in both cases we have a cubic polynomial. The only difference is this constant C. So these functions differ by a constant, because when we take the derivative, we get the quantity to X plus three squared, but the constant gets killed off. And so let's write that here. Let's put, um, that one and two only differ. I see, and that completes the problem.

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