Sketch the region enclosed by the given curves and find its area. y=√(x), y=(1)/(2) x, x=9

Name: Problem 25, Chapter 5: Calculus
Uploaded: 2021-07-20T18:14:49-08:00
Duration: 4 min 31 s
Channel: Sam Low
Description: Sketch the region enclosed by the given curves and find its area. y=√(x), y=(1)/(2) x, x=9

Sketch the region enclosed by the given curves and find its...

Key Concepts

Setting Up Integrals for Area Calculation

Once the region and its boundaries are determined, the next concept is to form an integral that expresses the area between the curves. This is done by integrating the difference between the upper and lower functions over the appropriate interval, based on the identified intersection points and any additional vertical or horizontal boundaries.

Definite Integration

After the integral is set up, definite integration is used to calculate the exact area of the region. This involves evaluating the integral with the proper limits of integration, thus determining the total enclosed area between the curves.

Graphical Analysis

Understanding the shape and layout of curves on a coordinate plane is essential. This involves sketching graphs of the functions to visualize the enclosed region, determine the area of interest, and identify the boundaries that will be used in the integration process.

Intersection Points

Finding where the curves intersect is a key step in determining the limits of integration. This typically involves setting the functions equal to each other and solving for the variable, which defines the interval over which the area between the curves is calculated.

Transcript

00:03 In this video, we're going to go over how to find the area in between two curves.

00:10 The area that it's requested is in blue.

00:14 And the first thing we need to do is find intersection points.

00:18 Now, it's pretty obvious that from the graph, one of them is zero.

00:22 And you can verify this by cross multiplying the two curves to each other, as you see.

00:30 Secondly, by setting 9 minus the square root of 9 minus x squared equal to the square of x squared plus you'll get the other intersection point that is x equals 2 and just doing a little bit of algebra there i verified this by just plugging in 2 into into the second line here and verified that 2 was a valid point negative 2 will satisfy this equation but it is not an intersection point that's because we're considering only positive x values now it's also apparent that x over the square root of x over the square root of x squared plus 1 is above x over the square root of 9 minus x squared.

01:24 So we can just subtract their integrals over the region from 0 to 2 and find that we're going to find the area between them that way.

01:37 Both of these integrals can be solved with a very basic u substitution.

01:42 For the first one, i let u be equal to x squared plus 1, and then i differentiated both sides.

01:49 Since du requires a 2x, i multiplied by 2, but at the same time, i divided by 2, just so that nothing would be lost.

02:00 I also had to change the bounds, and because i'm integrating something of the form 1 over 2 times square root of u, i know that because of my derivatives, if i integrate that, i'm going to get just the square root of u, which will be evaluated from 1 to 5, hence giving the square root of 5 minus 1.

02:21 Now the second integral is a little bit, is much the same, actually...

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