Sketch and find the area of the region bounded by the given curves. Choose the variable of integ

step 1 So so many students are used to different ways of writing problems. For instance, you might look

Sketch and find the area of the region bounded by the given...

Key Concepts

Sketching Curves

Drawing a sketch of the curves helps visualize the region of interest. It clarifies how the curves intersect and which parts of the region lie inside the boundaries, aiding in the proper identification of integration limits and the appropriate variable of integration.

Single Integral Setup

Setting up the area as a single integral means expressing the entire area calculation within one integration expression. This often involves manipulating the equations of the curves and the integration limits so that the area of the entire region can be obtained without breaking the integral into separate parts.

Region Bounded by Curves

This concept involves identifying the area confined between two or more curves. It requires understanding how these curves intersect and enclose a region in the plane, which is fundamental when setting up an integral to calculate the area.

Intersection Points

Determining the points at which the curves intersect is essential because these points define the limits of integration. They serve as the boundary values over which the area is integrated and ensure that the integral covers the entire region accurately.

Variable of Integration

Choosing the variable of integration is an important step when computing areas. Depending on the orientation and nature of the curves, integrating with respect to x or y can simplify the limits and the integrand, so selecting the optimal variable ensures that the problem is handled as a single, coherent integral.

Transcript

00:01 So so many students are used to different ways of writing problems.

00:06 For instance, you might look at this and say that's one third x equals y.

00:12 So pretty much both of these are equivalent.

00:16 So if it helps to write it this way, then write it that way that your slope is up one, write three.

00:22 And then the other function is this x equals 2 plus y squared.

00:33 So if you think of it as, i don't even know if this is helpful, but it's sort of like a quadratic, except it's tilted on its side.

00:45 So the y intercept being, well, x intercept, i guess i should say, is two.

00:52 And then it's slope like this.

00:57 I guess i should have made my point of intersection over here.

01:01 So it actually doesn't matter how you write.

01:03 Right the work.

01:06 But it's pretty evident that the point of intersection is at 3 -1.

01:12 And you can double -check by plugging in 3 -4 -x and 1 for y.

01:16 And then there's another point of intersection over here, which the slope is up 3, right, 1, that it's pretty evident that 6 -2 is another.

01:26 Now there's ways of figuring it out.

01:29 You know, you can set the equations equal to each other, like 3 -y is equal to 2 plus y squared and solve the problem this way y squared minus 3 y what's that plus 2 so factors of 2 that add to be negative 3 negative 2 and negative 2 and then the solutions are the opposite but then you still have to go back and find your x coordinates so probably the best way of doing this is leaving everything in terms of why because they gave you the problem that way so that's going from 1 to 2 and i showed you the work getting it that way.

02:05 Your upper function is the linear function, the 3y, and then make sure you subtract off the quantity of 2 plus y squared.

02:13 Because if you don't do the quantity, then you probably won't distribute that in there, and you'll get the wrong answer.

02:22 So you have to have negative y squared.

02:25 I like it in standard form, still from 1 to 2.

02:29 There's a lot of math to be done still.

02:32 So i hope i haven't simplified anything too far.

02:35 So we have negative one -third y -cubed plus three halves, y squared, minus two y from one to two, and then plug in your upper bound...