Problem

Evaluate the integral by interpreting it in terms…

01:52

Problem 38 Hard Difficulty

Evaluate the integral by interpreting it in terms of areas.

$ \displaystyle \int^5_{-5} \bigl( x - \sqrt{25 - x^2} \bigr) \, dx $

Answer

$-\frac{25 \pi}{2}$

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Video Transcript

Problem 38 here were asked to give an exact value for this integral in terms of area. So geometrically looking at the area of each of these functions. So think of this as you've got two functions, you've got to function y equals x. And then you have another function. Why equal 25 minus x squared square root. So let's look at these separately. So the function Y equals X. What does that look like? Well, you should know pretty well that is the line that goes straight through the origin. And so it's at the point -5 -5 5 five. And so if you think about it, the area here, so ah when you sum this up, this is gonna be what um negative 25. So sorry, it's gonna be 25 back up. Sorry, You look at negative areas under the axis. So five times five. So this is 25/2 with a negative sign. And then the area here is going to be five times 5, 25/2 with a positive sign. And that's just one half base times height. To figure out the area of these um triangles. So those are going to some out to zero. So this is going to be zero for the 1st part. Look at the next part here, square both sides, Y squared equals 25 minus x square. That means that X squared plus y squared equals 25. That is a circle centered at the origin. So a circle Centered at the origin with a radius of five. As And since I'm only looking at the positive values for why it happens to be a semicircle, so pretend that that is a nice semicircle. So that is a semicircle. So what is the area going to be there? Well, what's the area of a circle? The area of the circle is pi r squared area of a semicircle is pi r squared over two. So this area is going to be the radius is five, So this area is 25 pi over to. So if you are subtracting that value, This is -25 pi Over two, final answer, -25 pi over to. So that is the exact value of this definite integral. Just using the geometry of the functions that I see there.

Robert D.

Florida State University

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 2

The Definite Integral

Problem