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Evaluate the integral.

$ \displaystyle \int^{\pi}_{0} (5e^x + 3\sin x) \,dx $

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00:56

Frank Lin

00:37

Amrita Bhasin

Calculus 1 / AB

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Integration

Campbell University

Idaho State University

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.

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04:00

Evaluate the integral.…

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00:48

Evaluate the definite inte…

0:00

02:40

04:06

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08:29

That we are asked to do the integral from 0 to Pi of five E. To the X. Yeah. Plus sine of X. Uh So the anti directive, three sine of x. always like to double check I have the right thing before moving on. Um So the anti derivative, if you just remember that the derivative of E to the X. Is itself. Uh It makes sense that the anti derivative would be itself. And that constant just goes along for the ride. And you can double check that the director of the five G d f s five E to the X. So in the same uh like I think about how the derivative of sine is co sign, but then I think about how the derivative of co sign is negative sign. So it makes sense to me that this needs to be a negative, you know, the constant still goes along for the ride, but it needs to be there as you're doing this problem. And then you have your bounds from 0 to Pi. I mean you can double check that. This is correct because the derivative needs to equal uh the original problem. So now as you plug in your upper bound uh five E. To the pi minus three, co sign of pie. Uh And then I also always tell my students to make sure you plug in the lower bound as well. And this is especially important because uh well let's look at the unit circle real quick. So we draw the unit circle in green and Pie is over here At -10. So it's important because co sign His negative one there. So we're looking at five E. To the pie. So that would make this plus three E. to the zero power equals one. So five times one is 5 and then co sign of zero is right here. Where the x coordinates one, so three times one is just that value. Um So we're looking at five E. To the pie, 5 -3 is two, So 3 -2 gives me one and this is your final correct answer? Yeah, yeah.

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