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Evaluate the integral by interpreting it in terms of areas.

$ \displaystyle \int^3_{-4} \biggl| \frac{1}{2}x \biggr| \, dx $

$$\frac{25}{4}$$

01:34

Frank L.

00:33

Amrita B.

Calculus 1 / AB

Chapter 5

Integrals

Section 2

The Definite Integral

Integration

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

01:20

Evaluate the integral by i…

02:46

00:57

00:25

04:28

Compute the definite integ…

00:32

Evaluate the integral.

00:47

02:14

01:49

Evaluate the integrals.

00:41

Evaluate the integrals…

given the integral of the absolute value of one half X, we could go ahead and graft that and pull again the point when X is three. That would give us why is three halves over here and then plug in negative four. I would get us negative to absolute value is positive, too. And we want to find the area underneath. Each of these curves bounded between the curve and the X axis. And so to do that well, we have a triangle with a base of four. Height of two. So four times two is eight. Divided by two is four. So therefore, this triangle here has an area of four. Uh huh. And that's triangle here as an area of three times three halves, which was the nine halves. But then that divided by two. So let's go ahead and write that out. Nine hands. Dividing by two is the same thing is multiplying by 1/4. I mean, sorry, one half, and so that would end up being 9/4. So the area on the right side here is 9/4. Okay, Now we want to add these two together so we could just come up with common denominator. So we could change this for to 16 over four, and then we're finally ready to add the So let's do this. Cell circle needs to that we're gonna add in green. The area on the left side is four. Another way to say that is 16/4. The area on the right side is 9/4. Therefore, we end up with a total area of 25 fourth. Yeah, so that is our final evaluation for this integral using geometric area.

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