Find the area of the triangle with vertices at (-1,4) (2,-2), and (5,1) by integration.

Name: Problem 30, Chapter 5: Calculus
Uploaded: 2021-06-18T15:00:44-08:00
Duration: 8 min 17 s
Channel: Monica Miller
Description: Find the area of the triangle with vertices at (-1,4) (2,-2), and (5,1) by integration.

Find the area of the triangle with vertices at (-1,4)...

Key Concepts

Definite Integration for Area Calculation

This concept involves computing the area of a planar region by setting up and evaluating a definite integral. In calculus, the area under a curve (or between curves) is found by integrating a function that represents the boundary of the region. When applied to a geometric figure, such as a triangle, the region is described by functions or boundaries, and the area is determined by integrating across the relevant interval, effectively summing up infinitesimally small slices or strips of the region.

Setting Up Integration Bounds

Determining the appropriate limits of integration is a key step in using integration to calculate area. This involves analyzing the geometric region to identify the lower and upper limits along a chosen axis. For non-rectangular regions like a triangle, the bounds may vary along the integration variable, and careful consideration is required to correctly express these bounds, which may lead to splitting the integral into separate parts if the region's defining functions intersect.

Determining Equations of Lines

When the boundaries of the region are linear, it is necessary to find the equations of the lines that form the edges of the figure. These linear equations, typically expressed in slope-intercept or point-slope form, establish the relationships between the coordinates along the edges. This information is critical for setting up the integrals, as the equations provide the exact functions that serve as lower or upper limits in the integral used to compute the area.

Partitioning the Region

In some cases, the region whose area is to be calculated does not have a single set of integration limits due to changes in the boundary conditions. Partitioning the region into subregions, each with its own set of limits, allows for more straightforward integral setups. After computing the area of each subregion by integration, the total area is obtained by summing these individual areas, ensuring that the entire region is accurately accounted for.

Transcript

00:01 For this problem, we're given three points that when connected form a triangle, and we're asked to find the area of that triangle by integration.

00:11 So let's go ahead and plot these points.

00:15 So we have negative 1, 4, we have 2, negative 2, and then we have 5 .1.

00:23 Okay, so we can kind of see what that triangle looks like.

00:27 Now, if we have to do integration, it's always about top 3 .5 .5.

00:31 Function minus bottom function.

00:33 So our top function is going to be that same linear line that goes from our point negative one four over to our point five one.

00:43 But the bottom will change.

00:46 So the first thing we need to do is find all of these lines as functions.

00:52 So the best way to do that is to find our slope and then finer b value.

01:00 And then you can see how our top function minus bottom function changes in those regions.

01:06 Okay.

01:08 So first of all, let's consider from negative 1 comma 4 to 0 .5 .1.

01:16 So we're going to do change in y's over change in xes.

01:21 So that 5 minus a negative 1 will become a 6.

01:25 So we have negative 3 over 6, which is a nice negative 1 half.

01:31 So to create the equation, you can either use your point slope form or you can use your slope intercept form, but we know we're going to have a negative 1 half x.

01:41 And then, you know, we need to figure out what our other value is.

01:46 So let's use one of our points and actually put, let's put an x is negative 1 in and consider what would.

01:55 Our b value b so that we result in a value of four.

02:00 So one half plus b and equal four.

02:06 And then to solve for that b value, we're really subtracting a half from both sides.

02:11 So think of that four is an eight over two.

02:14 So when we subtract one half, we get seven over two.

02:18 Okay, we have our equation for our first line.

02:21 So let's get rid of some of that work.

02:23 And we're going to have to do the same thing.

02:25 For the other two.

02:30 So again, we are going to do our change in our ys over our change in our x's.

02:36 So we get a negative six divided by three, which gives us a slope of negative two.

02:45 Yay, not a fraction.

02:46 That makes it easier.

02:47 So now we're going to do the same thing.

02:50 Another thing to note about this, i know that my b value is positive because i can see that it crosses the y axis above zero.

02:58 So i just know to put a plus, but i just don't know what my b value is.

03:03 So we'll grab again one of our formulas, or one of our points.

03:08 We're going to put a negative one in and make sure that we get a four out.

03:12 So when we solve for b this time, we get a value of two.

03:19 And then one last one to do.

03:30 So here our change of wild.

03:32 Over our change of x ends up being 3 over 3 which is a value of 1 and i kind of check it i'm like okay i do have a positive slope that's good and then notice this if i continue this line it would definitely cross the y axis in the negative values so i'm just going to put it as minus but again we have to find it so we'll again use a point and this is a lot easier because it's just the x minus something.

04:05 So 5 minus what equals 1.

04:07 So we found that to be 4.

04:09 Okay, so we found all these equations, but we still have the whole integration to do.

04:15 So on the next page, i'm just going to bring over our picture here and our equations, and we're going to go ahead and do our integration...

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