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Evaluate the integral.$\int_{0}^{4}(4-t) \sqrt{t} d t$

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01:28

Frank Lin

00:37

Amrita Bhasin

Calculus 1 / AB

Chapter 5

Integrals

Section 3

The Fundamental Theorem of Calculus

Integration

Campbell University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

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 integral.…

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Evaluate the given integra…

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Find the integral.$\in…

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Evaluate.$$\int_{0}^{4…

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Evaluate integral

Alright, here's a fun integral. Um There's a lot of different kind of cleanup techniques which can help with. First we look and we think oh you stuff doesn't work as nice as I'd like and so you you need another technique. So there's one technique is called distribution. So basically I'm going to distribute the root tea to both parts That will give me four, can steer the one half minus T to the three halves, I'm converting to exponent form, so I can do reverse power roll. Yeah okay so that will give me then four times t to the three halves over three halves by reverse power roll minus 10 to the five halves over five halves. Um Then I can plug in my limits. So let's clean up just a little bit. This will be 8/3 cheated. The three has minus 2/5 T to the 5/2 and I still have to plug in my limits. So when I plug in four I get 8/3 4 to the 3/2ves minus 2, 54 to the five halves. And then the other terms will be zero when I plug in. So I'm done with that part. Now I'm just going to clean this up a little bit, this is eight thirds now four to the three halves, I can square root first for two and then go to the eight. So that's eight. Then for the four to the five halves I can square root to get two to the fifth is 32. Um Okay so I get 64/3 minus 64/5. And our final answer, if I clean that common denominator is 128 over 15. So that is then my solution to the integral anyway. Hopefully that helped have an amazing day.

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