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(a) determine a definite integral that will determine the area of the region shown and (b) evaluate this integral.

(a) $\int_{1}^{2} e^{-x} d x$(b) $e^{-1}-e^{-2}$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Campbell University

Harvey Mudd College

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.

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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(a) determine a definite i…

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I feel like it might be a type of here, but, um, even the negative X looks something like this. It's an exponential decay function. Um, and I think the intention is to go from X equals one two x equals to even though it's not clear, and we want the area under that curve. So for part A, our answer is going to be because it's area under the curve, the integral from 1 to 2 of that function. Either the negative X power D x Um, yeah, so that's all that's your answer for part A. Hopefully it makes sense. You do your lower bound, then your upper bound, and then your function the EPS. So now you're ready to do part B, which is the anti derivative of that. Now, to me, it makes perfect sense that the answer, because the derivative of E to the X is e to the X, that this is either the negative expo. When you consider the chain rule, you'd have to have a negative in front. If you don't believe me, try this. Take the derivative of this and all equal what we started with. So now we can do we can plug in our bounds now that we have the correct anti derivative. So we're looking at negative E to the negative second power minus negative. So I'm gonna change that two plus e to the negative first power. Um, now, I would leave my answer like this, but addition is communicative. So if you want to, you can switch the order around. Uh, it looks something like this. Uh, if you really want to get fancy for your answer to be, as you could rewrite each of those as fractions. So, like, one over E minus one over e squared. But where do you stop? One of these circumstances is fine. You could have stopped earlier. And don't forget about part A up here.

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