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Use calculus to find the area of the triangle with the given vertices.

$ (0 , 0) $ , $ (3 , 1) $ , $ (1 , 2) $

$\frac{5}{2}$

Calculus 2 / BC

Chapter 6

Applications of Integration

Section 1

Areas Between Curves

Campbell University

Harvey Mudd College

Baylor University

University of Nottingham

Lectures

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02:20

So we have these three points and they're going to form a triangle, and we want to find the area of the enclosed triangle. So let's just quickly plucked. Explains zero zeroes at the origin. 31 is around there, and 12 is up here. So we see that we have this triangle. If we connect these points and we want to determine the area of this triangle Ah, using calculus. So we're going to be integrating. Of course. We're going to need to know, uh, so going from over here to hear, we're going to need to know the top function. We're going to need to know this function. So this line and we're also going to want to know this line. So determining these equations of these lines is should be quite simple. Uh, because they're all in the form, like was an explosive. Be so here. The slope is too two acts, and we're touching the origin s. So that's that equation here. The slope is we're going from zero 23 So here this slope is 1/3 X and there's no be value. Since it's touching the origin here, the slope is negative. Half y equals negative half X. But of course, there is this be value, since it's not touching the origin. But I'll owe let you figure out how we get that plus meet. It works out to be plus five over too. Yeah, we saw for that because we know it's why equals negative Half X plus some B. And then we know that on this line we have, for example, we could take this 0.1, too, so we can just plug in the values two negative half one plus B. This lets us suffer B equals five over, too. That's how we get the the five over to here. Five over two. Okay, so now we have the equations of our lines. Uh, we have this one, this one and this one and then So what we need to figure out our, um, is the area. So let's do that area equals integral. So we're going to break up the integral here at X equals one. So the top line overhears two x and the bottom is 1/3 ex. So our first integral is going to be top two x minus 1/3 x on our limits of integration is We're going from X equals 02 X equals one. So we need to put those limits there. This is D X plus into go Now. We're going from one too. I believe it was three yet 123 over here. And our top function is negative. Half X plus five over too. So negative half X plus five over too. And then we're subtracting the bottom function, which is 1/3 ex. So we're going to subtract one over three x d. X now, before we carry out, these inter girls were just going to simplify them a little bit. Uh, integral 0 to 1 two x minus. Wondered X two is six over three. So we have five over three X, the ex plus integral from 1 to 3. Oh, here we have this five over, too, and then we have a negative 1/3 and then we have a negative half. So those workout too? Um, 1/3 plus 1/2. So to over six plus one over. Plus three over six. That works out to five over six x dx. Okay. And now let's just carry these intervals. So the anti derivative here is five over six X squared. We're going from 0 to 1. Then we add the anti derivative. Here is five over to X minus five over 12 X squared. And this is going from 1 to 3. Now we just plug in our numbers. So here we have five over six, minus zero plus plugging in three. We have three times five is 15 over too. Minus plugging in three again. We have 88 times five is 40 40 over 12 and then we plug in one. So five over too. Um, minus five over 12. So after we simplify all of this, we end up with the value of five over too. So five over two is the enclosed area.

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