Quick question about total area (involving integrals)

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Homework Help Overview

The discussion revolves around finding the total area between the curve y=x^2-4 and the interval [0,3], specifically addressing how to set the upper and lower limits for integration when the curve crosses the x-axis.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore how to break the integral into parts based on where the curve intersects the x-axis, questioning how to determine the appropriate limits for integration.

Discussion Status

Some participants have provided guidance on the need to identify where the curve crosses the x-axis to correctly set the limits for integration. There are multiple interpretations of how to approach the problem, with suggestions on different ways to express the integral.

Contextual Notes

There is a focus on understanding the implications of the curve being above or below the x-axis and how this affects the calculation of area. The original poster expresses familiarity with the integration process but seeks clarity on limit determination.

Aerosion
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Homework Statement



Find the total area between 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 \int_{2}^{3} x^2 dx - \int_{0}^{2} 4 dx, 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.
 
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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.
 
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
 
Aerosion said:

Homework Statement



Find the total area between 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 \int_{2}^{3} x^2 dx - \int_{0}^{2} 4 dx, 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
-\int_2^3 (x^2- 4)dx
so the "total area" between the curve and the x-axis is
\int_0^2 (x^2- 4)dx- \int_2^3 (x^2- 4)dx



Homework Equations





The Attempt at a Solution

 

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