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Using the result of Exercise $31,$ evaluate (a) $\int e^{2 x} d x,$ (b) $\int e^{-x} d x$ (c) $\int \frac{1}{e^{4 x}} d x.$

(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

Campbell University

Harvey Mudd College

Baylor 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.

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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All right. So we're using this result that improved in in, uh, this problem that the anti derivative of E to the X DX is he get the one over a e to the X. So when we look at a problem like part A, which is asking you for e to the two x d. X. Basically, what it boils down to is everywhere that there's an A in this problem you replaced with to see how to is replacing the A and that problem. So replace this A with two, and this a with two and everything else is the same. So there's your first answer for part A. As you look at part B, where now it's negative. X. It's worth the while to recognize that, uh, this problem e to the negative. X is the same thing as negative. One X is it's okay to put a one in their negative X is the same as negative one times X eso won divided. A negative one is just negative one where you don't have to write negative one, just like it wasn't there to begin with. And then again, a is negative one plus C and before moving on The very last one is ah e to the who is kind of complicated. Let's do this in a different color because you have to think a little bit the inter rule of one over E to the four x t. X. So before doing this problem, what I would do is think about, uh, that that is the same thing as being in the denominators and negative exponents. So now replace a in that problem, this A with negative four. So this a need to replace the neck for this? They need to be replaced. Negative four. So we go back to green, so one over negative four e to the negative four X. Let's see now. I would let my students leave their answer like this, but you might also prefer writing the answer back with Positive Expo in. So, like negative 1/4 e to the four X, Um, yeah, plus Sisto. So this is a good answer for CIA's. Well, so there you have it

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