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

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

More Answers

02:41

Frank L.

01:35

Amrita B.

Topics

Integrals

Integration

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 2

The Definite Integral

Discussion

You must be signed in to discuss.

Top Calculus 1 / AB Educators

Kayleah T.

Harvey Mudd College

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor

Joseph L.

Boston College

Watch More Solved Questions in Chapter 5

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

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

Topics

Integrals

Integration

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 2

The Definite Integral

Top Calculus 1 / AB Educators

Kayleah T.

Harvey Mudd College

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor

Joseph L.

Boston College