Calculus I - Area Between Curves - Mistake on Answer Key

In summary, the conversation discusses the proper way to integrate and determine the area under a curve. It is clarified that integrating from 0 to 25 or from 25 to 0 will produce the same positive area, contrary to previous belief. The reasoning behind this is explained and it is confirmed that the correct way to integrate is from the lower value to the higher value. The use of vertical and horizontal strips is also mentioned.
  • #1
GreenPrint
1,196
0
This is the answer key to one of my quizzes

If you notice in the question, see attachment, were it says to integrate with respect to y the integral is integrating from 0 to 25 but this produces a negative area so this is technically wrong, yes? You don't just simply integrate from the lower value to the upper value for all cases? Like in this problem the proper integral would of been from 25 to 0 were your actually integrating from the higher value to the lower value... the reasoning behind this is because the higher valued limit appear to the left of the lower value limit 0 so the x coordinate of the higher value limit is -5 which is less than the x coordinate of the lower valued limit 0

the points of intersection of the functions are:
(-5,25)(0,0)
so is my reasoning correct for as to why you actually would integrate from 25 to 0 instead of 0 to 25 to produce a positive area?
 

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  • #2
GreenPrint said:
This is the answer key to one of my quizzes

If you notice in the question, see attachment, were it says to integrate with respect to y the integral is integrating from 0 to 25 but this produces a negative area so this is technically wrong, yes?
No. Integrating from 0 to 25 produces a positive number, 125/6, the same as the other integral.
GreenPrint said:
You don't just simply integrate from the lower value to the upper value for all cases? Like in this problem the proper integral would of been from 25 to 0 were your actually integrating from the higher value to the lower value... the reasoning behind this is because the higher valued limit appear to the left of the lower value limit 0 so the x coordinate of the higher value limit is -5 which is less than the x coordinate of the lower valued limit 0

the points of intersection of the functions are:
(-5,25)(0,0)
so is my reasoning correct for as to why you actually would integrate from 25 to 0 instead of 0 to 25 to produce a positive area?
 
  • #3
I got a positive area doing it the way it's written on the answer key.
 
  • #4
Hm I guess I entered it into my calculator wrong thanks, so it's always the integral from the lower value to the higher value after all than?
 
  • #5
Yes, if you are using vertical strips. The area of one of these vertical strips is (<upper y value> - <lower y value>) * dx.

If you're using horizontal strips, the area of a strip is (<right x value> - <left x value>) * dy.
 

1. What is the concept of "area between curves" in Calculus I?

The concept of "area between curves" refers to the area enclosed by two curves on a given interval. This can be calculated using integrals in Calculus I.

2. How can a mistake on the answer key affect the calculation of the area between curves?

A mistake on the answer key can significantly alter the calculated area between curves. This can lead to incorrect solutions and a misunderstanding of the concept.

3. What are some common mistakes made when calculating the area between curves in Calculus I?

Some common mistakes include incorrect integration limits, forgetting to include absolute value signs when necessary, and making errors in the integration process.

4. How can students avoid making mistakes when calculating the area between curves in Calculus I?

Students can avoid making mistakes by carefully checking their work and double-checking their integration steps. It is also helpful to practice and get feedback from the instructor.

5. How important is it to catch and correct mistakes in the area between curves in Calculus I?

Catching and correcting mistakes in the area between curves is crucial in Calculus I. This concept is fundamental to understanding more complex topics in calculus, and mistakes can lead to incorrect solutions and a lack of understanding of the material.

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