# Help explain curves which are entirely below the x-axis

#### JoKo

Can someone help explain to me or maybe show me how to find the area of a curve which is below the x-axis. If the answer is a negative, is it correct ?

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#### studiot

Can someone help explain to me or maybe show me how to find the area of a curve which is below the x-axis. If the answer is a negative, is it correct ?
First let's correct your terminology.

Curves don't have areas.

They have length, curvature and slope.

An area can be defined between a curve and another line.

The simplest area is if that other line is one of the axes.

We call the area between the x axis and the curve, "The area under the curve"

Since this question is asked in calculus, I can say that this area corresponds to
the definite integral between points (limits) on the same side of the axis as each other.

Integrals below the x axis are counted (work out) as negative

Integrals above the x axis are counted (work out) as positive

BUT Areas are always positive

For this reason the definite integral may be different from the total area under the curve.

So the integral of a sine curve from 0 to 360 is zero,

but the area under a sine curve is twice the integral of the sine from 0 to 180.

http://www.bbc.co.uk/education/guides/zq3ggk7/revision/1

Does his help ?

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#### skipjack

Forum Staff
I doubt that JoKo meant the area of a curve. The area bounded by various curves or lines was probably intended. Such an area can't be negative, regardless of whether or not the x-axis is part of the boundary.

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#### Country Boy

Math Team
For example: find the area of the region bounded by the x-axis, y= 0, and the parabola, $$\displaystyle y= x^2- 1$$.

The first thing you should do is draw the graph- or at least imagine it- to see that the two cross at x= -1 and x= 1, that the only region bounded by those is between x= -1 and x= 1, and that, in that region, the x-axis is always above the parabola.

To find the area, subtract the equation of the lower curve from the equation of the higher curve, 0- (x^2- 1)= 1- x^2, and integrate from x= -1 to x= 1:

$$\displaystyle \int_{-1}^1 1- x^2 dx=$$$$\displaystyle \left[ x- \frac{x^3}{3}\right]_{-1}^1=$$$$\displaystyle (1- \frac{1}{3})- (-1+ \frac{1}{3})$$$$\displaystyle = 2- \frac{2}{3}= \frac{4}{3}$$.

Of course, "area" of any region is a positive number and we guaranteed that by subtracting "the equation of the lower curve from the equation of the higher curve".

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#### JeffM1

I have thanked so many posts here because I have learned something from this interesting question.

I must admit that I had always interpreted the "area under the curve" as meaning a number that was positive if the area was above the x-axis and negative if below the x-axis. That is, I was using the term in a figurative or perhaps conventional sense rather than a literal sense. Obviously an area must be positive, but a limit can be positive, negative, or zero.

So thanks to everybody.