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Evaluate the given integral and check your answer.$$\int\left(2 e^{t}-3 t^{4}+\frac{7}{t}-12\right) d t$$

$$2 e^{t}-\frac{3 t^{5}}{5}+7 \ln |t|-12 t+c$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 1

Anti differentiation - Integration

Integrals

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

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. Our test is to do the integral of two e to the T power minus three t to the fourth, uh, plus seven over tea minus 12 d t on. There's two things in this problem. The first is that the derivative of I'll you see times natural Log of t is equal to that constant divided by t. That's something we've learned before. So if we're doing the integral, we've got to go backwards. So that's for this part of the problem, or C would be seven on then. The second part is you know, if we have d d t of some constant times e to the T, that derivative is also k times e to the t. So we're going backwards. It's gonna be the exact same thing. So that's that part. So we're already ready to find the anti derivative. Well, as I mentioned, since the derivative of two e to the T is to e to the T that going backwards is also to e to the T. Whereas the next one. This is where we're adding one to the exponents. Four plus one is five and then you divide by a new explosion so that three is still there, but divided by five Now. Now the next piece is from right here, where the our final answer will be seven Natural log. Now we do wanna put an absolute value in here because in this problem it's assumed that t is positive. Whereas over here it's not assumed we can Onley log positive numbers. So that's why we have that. And then the derivative of negative 12 t is negative. 12. I just think of it that way and then don't forget about a constant plus C and you can double check that This answer is correct that I'm circling in green by taking the derivative of that and you'll get the that answer that interval eso we should be good.

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