# Quick question about total area (involving integrals)

## Homework Statement

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

## The Attempt at a Solution

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.

## The Attempt at a Solution

Dick
Homework Helper
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.

Gib Z
Homework Helper
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
$$\int^3_0 x^2-4 dx = \int^3_2 x^2 - 4 dx + \int^2_0 x^2 - 4 dx$$

HallsofIvy
Homework Helper

## 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]?

## The Attempt at a Solution

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
$$\int_0^2 (x^2- 4)dx- \int_2^3 (x^2- 4)dx$$