Find the area beween the curves y=x^2 and x+y=2 and the x axis

  • Context: MHB 
  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Area Axis Curves
Click For Summary
SUMMARY

The discussion focuses on calculating the area between the curves defined by the equations \(y=x^2\) and \(x+y=2\), as well as the x-axis. The intersection points of these curves are identified at \(x=-2\) and \(x=1\). To find the area between the curves, the larger area is calculated using the integral \(\int_{-2}^{1} (-x^2 - x + 2)\,dx\), resulting in an area of \(\frac{9}{2}\). The participants clarify the setup for the area calculation, confirming that the correct approach involves integrating the respective functions over the specified intervals.

PREREQUISITES
  • Understanding of integral calculus, specifically definite integrals.
  • Familiarity with curve intersection and area calculation between curves.
  • Knowledge of the properties of quadratic functions and linear equations.
  • Ability to interpret and manipulate mathematical expressions and integrals.
NEXT STEPS
  • Study the method of finding areas between curves using definite integrals.
  • Learn how to determine intersection points of functions algebraically.
  • Explore the application of the Fundamental Theorem of Calculus in area calculations.
  • Practice integrating various functions to solidify understanding of area under curves.
USEFUL FOR

Students and educators in calculus, mathematicians interested in area calculations, and anyone looking to deepen their understanding of integration techniques and curve analysis.

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
Find the area beween the curves $y=x^2$ and $x+y=2$ and the x axis

First on graphing these the $x-axis$ seem irrelevant in that it is outside the area to find.

[desmos="-10,10,-10,10"]y=x^2;y=-x+2[/desmos]
 
Physics news on Phys.org
How would we find the area between each curve and the x-axis individually?

Where do these curves intersect?
 
Set the equations equal to each other.
$x=-2$ $x=1$
 
karush said:
Set the equations equal to each other.
$x=-2$ $x=1$

Good! Ok, so now to find the area between them we need to find the area of the bigger part and then subtract out the area of of the smaller part. Just like if we had a whole pizza with a piece missing, we could take the area of the whole pizza minus the area of one slice to find the area of the whole pizza with a piece missing.

In this problem, which function has a larger area or put another way, which function is larger than the other over the interval?
 
$\int_{-2 }^{1 } (-x^2 - x+2)\,dx$
$\left[-\frac{{x}^{3}}{2 }-\frac{x}{2}+2x \right]^1_{-2}=\frac{9}{2}$

I hope
 
I am thinking, given the original problem statement, that you are to find the following area:

View attachment 4411
 

Attachments

  • karush3.png
    karush3.png
    2.6 KB · Views: 138
For the interval $0\le x\le2$ I get $-\frac{2}{3}$
 
Did you set it up as:

$$A=\int_0^1 x^2\,dx+\int_1^2 2-x\,dx$$ ?
 
No that would be 2/3
 
  • #10
I get a different result, but I can't tell where you and I differ without seeing your work...:D
 
  • #11
Yes, you are correct Mark. Sorry for misreading the question karush. I was thinking of this region in light green but Mark is correct. :)

[desmos="-10,10,-10,10"]y<x^2\left\{-2<x<1\right\};x+y<2\left\{-2<x<1\right\};[/desmos]
 
  • #12
No problem, how did you get the shading in demos?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
877
  • · Replies 17 ·
Replies
17
Views
3K