Quick question about total area (involving integrals)

In summary: Okay, so I already know the answer, but I'm unsure about one thing. If you want the sum of the two instead of the difference then you need to figure out where the curve crosses the x-axis. And you can't break up the integral like you did. You can either do \int^3_0 x^2 - 4 dx= \int^3_0 x^2 dx - \int^3_0 4 dx or\int^3_0 x^2-4 dx = \int^3_2 x^2 - 4 dx + \int^2_0 x^2 - 4 dx
  • #1
Aerosion
53
0

Homework Statement



Find the total area betwen the curve y=x^2-4 and the interval [0,3].

Homework Equations





The Attempt at a Solution



Okay, so I already know the answer, but I'm unsure about one thing.

It's ture that you have to break the equation such that it says [tex]\int_{2}^{3} x^2 dx[/tex] - [tex]\int_{0}^{2} 4 dx[/tex], but how would you go about finding which upper and lower limits to set for the two integration problems like that?

I know how to take care of the rest once I find those, but I'm not sure how to get the upper and lower limits once I break the equation into two.
 
Physics news on Phys.org
  • #2
The integral will give you the area above the x-axis minus the area below the x-axis. If you want the sum of the two instead of the difference then you need to figure out where the curve crosses the x-axis.
 
  • #3
And you can't break up the integral like you did. You can either do
[tex]\int^3_0 x^2 - 4 dx= \int^3_0 x^2 dx - \int^3_0 4 dx[/tex] or
[tex]\int^3_0 x^2-4 dx = \int^3_2 x^2 - 4 dx + \int^2_0 x^2 - 4 dx[/tex]
 
  • #4
Aerosion said:

Homework Statement



Find the total area betwen the curve y=x^2-4 and the interval [0,3].
Do you mean "find the area between the curve y= x2- 4 and the x-axis for x in [0,3]?

Homework Equations





The Attempt at a Solution



Okay, so I already know the answer, but I'm unsure about one thing.

It's ture that you have to break the equation such that it says [tex]\int_{2}^{3} x^2 dx[/tex] - [tex]\int_{0}^{2} 4 dx[/tex], but how would you go about finding which upper and lower limits to set for the two integration problems like that?

I know how to take care of the rest once I find those, but I'm not sure how to get the upper and lower limits once I break the equation into two.

Homework Statement

No. y= x2- 4 crosses the x-axis at x= 2. For x> 2, y is negative and the integral is the negative of the area. The area between the curve and the x-axis for x between 2 and 3 is
[tex]-\int_2^3 (x^2- 4)dx[/tex]
so the "total area" between the curve and the x-axis is
[tex]\int_0^2 (x^2- 4)dx- \int_2^3 (x^2- 4)dx[/tex]



Homework Equations





The Attempt at a Solution

 

Related to Quick question about total area (involving integrals)

1. What is the formula for calculating total area using integrals?

The formula for calculating total area using integrals is ∫f(x)dx, where f(x) represents the function that describes the shape or area being calculated and dx represents the infinitesimal width of each rectangle used to approximate the area.

2. Can integrals be used to calculate the area of irregular shapes?

Yes, integrals can be used to calculate the area of irregular shapes. By breaking the shape into infinitesimally small rectangles, the total area can be approximated and then calculated using the integral formula.

3. How do the limits of integration affect the calculation of total area?

The limits of integration determine the range of values that are used to calculate the integral. These limits can be adjusted to only include the desired area and exclude any unwanted portions, thus affecting the total area calculation.

4. Can integrals be used to calculate the area under a curve?

Yes, integrals can be used to calculate the area under a curve. This is known as the definite integral and is represented by the notation ∫f(x)dx from a specific lower limit to a specific upper limit.

5. What are some real-world applications of using integrals to calculate total area?

Integrals are commonly used in physics, engineering, and economics to calculate the area under a curve, which represents the displacement, velocity, or acceleration of a moving object. They are also used in calculating the area of irregular shapes in architecture and land surveying.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
2K
  • Calculus and Beyond Homework Help
Replies
10
Views
484
  • Calculus and Beyond Homework Help
Replies
23
Views
979
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
567
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
964
  • Calculus and Beyond Homework Help
Replies
6
Views
971
Back
Top