Problem

Find the area of the region bounded by the parabo…

14:16

Question

Answered step-by-step

Problem 55 Easy Difficulty

The curve with equation $ y^2 = x^2 (x + 3) $ is called Tschirnhausen's cubic. If you graph this curve you will see that part of the curve forms a loop. Find the area enclosed by the loop.

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Madi Sousa

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 6

Applications of Integration

Section 1

Areas Between Curves

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Watch More Solved Questions in Chapter 6

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

Video Transcript

in this problem, we have to use integration, define the area under a curve. And so we have got to talk about how we do that for the function that were given. We're given. Our function is why squared equals X squared Times X plus three. So we can simplify that a little bit and say that why is equivalent to plus or minus the square root of X square Times X plus three. So when we graft that this is a very rough drawing. But this is what the graph would mostly look like. We have an intersection at 00 and if you drew this better than I did, you would see that this is symmetric and so we can do a little bit more simplification with our function. We can say that why equals the absolute value of X Times Square root of X plus three. But working with absolute values and intervals is a little bit more challenging than what we need here. So we can say when X is less than zero. Why equals negative X time to square root of X plus three And now remember when we look at this graph, or maybe you put it into a graphing, um, utility to look make it look better than my drawing you find in its symmetric. So what you could basically do. I find the area of the top portion or the bottom portion off our function and multiply it by two. And that's what we're going to do. We'll gather. The area is equivalent to two times the integral of negative 3 to 0 of negative X times the square root of X plus three with respect to X. So now we can do some U substitution. We can say let you be X plus three. So U minus three is equal to X and d use equal to D X, and now we can change. The limits of integration are a value which was negative. Three. We can add three and we'll get zero for a B value. Well, we had zero before. We'll add three and we'll get three. So our new integral now using new substitution will look like this. Are area will be equivalent to negative two times the integral from 0 to 3 of U minus three times the square root of you in do you and now we can simplify that and say that our area equals negative. Two times the integral from 0 to 3 of you raised the three halves minus three times you raise to the half power and do you? So now we can take the anti derivatives will get negative two times two or five. You raised a five over to minus three times. Pardon me, That should know. Sorry to Over three. You raised a three over to when will evaluate this from 0 to 3. Sorry for the confusion. I just cause they're hopefully it becomes clear in this next step. So now we'll basically plug in. Our limits of integration will get negative two times. Two over. Five times three raised to four over. Two times three raised to one half, minus two times three raised to two over. Two times three. Race to the one half. Now I know that these could be simplified, which we're going to do when we simplify. This whole expression will get negative two times 2/5 times nine times a spirit of three minus two times three squared of three. Now we can put all of this in a common denominator and we'll get negative 36 plus 60/5 times square root of three. And finally, what gets the area of our curve is 24/5 times the square root of three. So I hope that this shows you how we can use integration to find the area under a curve. Specifically, we looked at a symmetric function and how we can go through those steps to find the area using you substitution as well.

Get More Help with this Textbook

James Stewart

Calculus: Early Transcendentals

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

Problem

Find the area of the region bounded by the parabo…

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Grace He

Catherine Ross

Caleb Elmore

Kristen Karbon

Calculus 2 / BC Courses

Recommended Videos

Watch More Solved Questions in Chapter 6

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Caleb Elmore

Kristen Karbon

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Catherine Ross

Caleb Elmore

Kristen Karbon

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Caleb Elmore

Kristen Karbon

James Stewart

Calculus: Early Transcendentals