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Evaluate the integral.

$ \displaystyle \int^{2}_{0} (2x - 3)(4x^2 + 1) \,dx $

$-2$

01:21

Frank L.

01:15

Amrita B.

Calculus 1 / AB

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Integration

Missouri State University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Lectures

03:09

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31:55

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01:48

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08:52

we need to do the integral from 0 to 2. Uh This product of by no meals. Uh So what that means is we don't have the product rule for integral. So what we need to do is make sure that we distribute everything out four X squared plus one. Yeah before taking each individual anti drop so that's exactly what I'm doing. Um So we have the integral from 0-2. Just rewriting it of eight x cubed. I would actually just write it this way just out of habit 12 X squared and then the two X. And just double check that my arithmetic is correct. Um Because if I keep working I might have the wrong answer at the end. Um From a simple mistake but anyway the anti gravity add one to the exponent. You divide by your new exponents. So eight divided by force. To add one to the exponents, divide 12 to 53 is four, add on to the exponent to divide by two is one and then minus three X. And you can double check. That's correct by um The derivative of what's in blue needs to be the original problem so now I can start plugging in my upper bounds so I'm plugging in to here here, here and here. Um So two to the 4th, 248 16, That's multiplied by two times. The other two would be 32 To the 3rd of the eight, two times two times two. But then temps four is 32 two squared is 43 times to would be six. So pretty clearly these two things cancel out. Um And I always encourage my students to plug in zeros in here, even though in this stage it's kind of obvious that you're going to get a bunch of zeros in the future, it won't always be that case. But anyway you're going to subtract off zero there As you look at the last remaining two pieces for -2 is equal to negative to the correct answer.

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