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# Use calculus to find the area of the triangle with the given vertices.$(2 , 0)$ , $(0 , 2)$ , $(-1 , 1)$

## 2

#### Topics

Applications of Integration

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

So we have these three points which are going to form a triangle, and we want to determine the area of the enclosed triangle. So let's put the points 20 02 minus 11 No, we collect, We connect them to find the triangle and we want to determine the area of this enclosed region here. To do that, we're going to need the equations of the line of the lines. So this one over here is why equals the slope is one. So why equals X and the Y intercept is too. This bottom one here, why is evil to the slope? Here is negative one so negative x and again the Y intercept is too. And this last one is why is equal to the slope? Here is negative one over three x plus some B values. So let's figure out to be valued. We know why equals negative 1/3 x plus be We can plug in any point on this line. So, for example, we have this 0.20 so we can just plug in zero equals negative one over three times two plus b and then what we get is 2/3 equals B. Okay, so negative. 1/3 X plus two over three. Great Know what we want to do is integrate. To figure out the area area is equal to integral. So first word we're going from X equals negative. One toe X equals zero. It's going to be top function minus bottom function. The top function is X plus two X plus two minus the bottom function, which is negative. 1/3 X plus two or three negative 1/3 X plus two over three d X plus. Now we're integrating from zero two x equals two. It's going to be the top function, minus the bottom function. Top function is negative. X plus two. Uh, over here, that's negative. X Plus two negative X plus two and then we're subtracting the bottom function, Which again is the line negative. 1/3 X plus two over three d x. Okay, once we calculate, all of this will have the area. So let's just simplify things a little bit First. Integral Negative 1 to 0. Here we have an okay, let's use a different color. Here we have X plus, wondered X. So that works out the 4/3 x then we have a plus two and then we have a minus 2/3 s. So that works out to plus four over three DX plus integral from 0 to 2. Let's simplify this a little bit. We have a negative X plus a 1/3 ex that works out to negative 2/3 X. Next, we have a plus to minus 2/3 which is plus for over three. And in all of this DX. So now we just carry out the intercourse. The anti derivative of the first integral is for over six x squared, plus four over three acts going from negative 1 to 0. That's zero plus, the anti derivative here isn't negative to over six X squared plus four over three tonnes X and this is going from 0 to 2. Now we just plug in the numbers so plugging and zero here just gives us a zero No plugging in a minus one. Here we have a four over six, which is 2/3 uh, plus for over three times minus one. Next we plug in to two. Squared is four. That's an eight over six. Ah, plus, he re plug in to hear we have eight over three, and then we subtract and we plug in zero, which gives us a zero. Okay, so the zeros air gone. And now, after we simplify everything we end up with. So this 1st 1 is going to give us A to over three. And we have also plus four over three, which works out to six over three, which is to so the area then closed region is too.

University of Toronto

#### Topics

Applications of Integration

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp