Calculating the area of regions between two curves

[Saba]

Homework Statement

When calculating the area of regions between two curves, why do we igone that they have crossed x-axis, and simply use the formula of (definite integral of upper function - definit integral of lower function)? dosen't the area under x-axis region become negative?
An example has been attached, thank you.

Relevant Equations

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[DrClaude]

You want the area in the physical sense of the word, i.e., a positive quantity. You will calculate f(x)-g(x), which will be bigger if g(x) < 0 as the two negative signs will make a plus. As you can see in the figure, this is what you want because in the region where g(x) is below zero the area is greater than the area delimited by f(x) down to y = 0.

 

[Saba]

Ohh so essentially the definit inetgral wouldn't take the diffrence between the area under x-axis and the area above x-axis?

 

[DrClaude]

Ohh so essentially the definit inetgral wouldn't take the diffrence between the area under x-axis and the area above x-axis?

I'm not sure I understan what you mean here.

I think that you have to be careful about the use of the expression "area under the curve" to explain the integral. If you take
$$
\int_0^{2 \pi} \sin x \, dx = 0
$$
you get zero because the integral over $(0,\pi)$ cancels out the integral over $(\pi,2 \pi)$. But if you are interested in the (surface) area between the sine curve and the x-axis, i.e., between $f(x) = sin(x)$ and $g(x) = 0$,

then you have to consider that two functions cross at $x = \pi$ and the result will be
$$
\int_0^{\pi} (\sin x - 0) \, dx + \int_{\pi}^{2 \pi} (0 - \sin x) \, dx = 4
$$

 

[FactChecker]

Ohh so essentially the definit inetgral wouldn't take the diffrence between the area under x-axis and the area above x-axis?

The definite integral between the curves y=0 to y=f(x) DOES distinguish between above versus below the x-axis. It will give you the area above minus the area below. But a definite integral between the curves y=g(x) to y=f(x) only distinguishes between f(x) being above g(x) versus f(x) being below g(x). So that is the only thing you need to worry about in the original post.

 

[Mark44]

I don't think this has been mentioned. Given your drawing in post #1, the total area, $A_1 + A_2$ (if that's what you're after) will require two integrals.

The first integral, whose value is $A_1$, is $\int_{-3}^0 f(x) - g(x)\,dx$. The second integral, whose value is $A_2$, is $\int_0^{2?} f(x) - g(x)\,dx$ for the reason that on that interval $f(x) \ge g(x)$. I've estimated the right-hand endpoint of that interval, since your drawing doesn't show it.