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Compute $\frac{d}{d x}\left(\frac{1}{a} e^{a x}\right),$ where the constant $a \neq 0 .$ Use this result to prove that $\int e^{a x} d x=\frac{1}{a} e^{a x}+C.$

(a) $1 / 2 e^{2 x}+c$(b) $-e^{-x}+c$(c) $-\frac{1}{4} e^{-4 x}+c$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 1

Anti differentiation - Integration

Integrals

Missouri State University

Campbell University

Idaho State University

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.

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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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Calculate.$$\int \frac…

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Use integration by parts t…

right. So this problem basically walks you through it. Uh, like they tell you to find the derivative of one over a e to the a X. And if you follow the rule for the product rule Uh, sorry. The chain rule. I don't know where my mind's at. What you do is you take the derivative of Well, first of all, leave one over a alone that z just the, um a constant. The derivative of E to the A X is e to the a X, but then you do the channel, which is taking the derivative of a X Well, since a is a constant, a derivative of X is just 18 times one would give me a now Before stopping this, I want to point out that the one over a times a would cancel each other out is your dividing by a and later multiplied by a. So this derivative, simply put, is e to the X. So what that means is, when I do the anti derivative of e to the A X or the integral of E to the A X dx, it's going backwards. So the answer to this is one over a e to the X And don't forget about your plus seat. Eso That's what that means. Since the derivative of this is this then the anti derivative of this is this There you have it.

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