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Evaluate the given integral and check your answer.$$\int \frac{5 t^{7}-2 t^{4}+3}{2 t^{3}} d t$$

$$\frac{2 t^{7}-2 x^{4}-12}{4 t^{2}}+c$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 1

Anti differentiation - Integration

Integrals

Campbell University

Oregon State 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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Evaluate the given integra…

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Evaluate the integral.…

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Evaluate the integral.

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Evaluate the integral and …

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I always make my students rewrite this problem because we're dividing by a mono mule to take cubed is a mono mule. Uh, d t. And, uh, what I make my students do is, well, that's not what that number waas was. Five t to the seventh, minus two t to the fourth plus three. So for whatever reason, uh, several students are good with if we had three separate fractions and getting the same denominator to combine them together But they don't realize that you can split up a Zara's. It's a mono me on the bottom into three separate fractions. The reason why I would do that just hit, equals and rewrite it is now we have five have five, divide by two and you can subtract your exponents. We're looking at T to the fourth Power seven minus three is four minus two. Divide by two is of one. I don't have to write down one. If I subtract these exponents, I get t two the first power and then plus three halves t to the negative Third. If it helps, you know, you might want to write down that this is like t +200 minus three is negative. Three a d. T. So now we can do the anti derivative, which is adding one to the exponents. Four plus one is five. If I divide by five, they'll actually cancel out this five. So we're looking at one half a T to the fifth minus. Add one to your exponents, multiply by the reciprocal of that new exponents. Um, now, be careful because you're adding one Thio negative three. So it's negative, too. And when I divide by negative two is gonna be stuck into the denominator with this one eso it becomes three force notice that the sign did change and then don't forget about plus C now, in my experience, most teachers will leave your answer like this. This should be good enough. You can check your work by taking the derivative to get you back up here. That's how you can check your work that you have the right answer was a terrible arrow. Um, but you can also, if you really want to, you could get the same denominator. Um, and the denominator would be based off of that for t to the negative second power to get it back into the denominator. So then you just have to methodically think about Oh, well, this is one half with the denominator four would be to force. And to make a t to the fifth, you'd have to have seven. Because seven minus two is five. And this will be another minus two and b t to the fourth, because two forces, one half four minutes to is to And then I think that fraction is good. So B minus three and then plus C, But this is, uh this is like algebra I wouldn't classify. This is calculus. I'd rather leave my answer. Is this one up here?

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