Negative area above x-axis from integrating x^2?

Suppose the following integration,

##\int_3^{-1} x^2 \, dx = \frac{1}{3}(-1)^3 - \frac{1}{3}(3)^3 = -\frac{28}{3}##

However, if we have a look at the graph,

The area between ##x = 3## and ##x = -1## is above the x-axis so should be positive. Dose anybody please know why the I am getting negative area from the integral?

Many thanks!

Integrals are oriented areas. ##\int_a^b f(x)\,dx =-\int_b^a f(x)\,dx.## So it has an opposite sign if you integrate from left to right or from right to left.

fresh_42 said:

Integrals are oriented areas. ##\int_a^b f(x)\,dx =-\int_b^a f(x)\,dx.## So it has an opposite sign if you integrate from left to right or from right to left.

Thank you for your reply @fresh_42 !

That is very helpful! Does this mean since the integral is negative, then the area is negative, so we can have negative area above the x-axis (without going below the x-axis)?

Many thanks!

ChiralSuperfields said:

Suppose the following integration, ##\int_3^{-1} x^2 \, dx = \frac{1}{3}(-1)^3 - \frac{1}{3}(3)^3 = -\frac{28}{3}##

ChiralSuperfields said:

Does this mean since the integral is negative, then the area is negative, so we can have negative area above the x-axis (without going below the x-axis)?

No. The left-most integral above has the limits of integration in the wrong order, so the value of the integral will be negative, as you found. If you rewrite the limits in the proper order -- left to right or bottom to top, then you'll get a positive value.

Your integral should have its limits like so: ##\int_{-1}^3 x^2 \, dx##.

ChiralSuperfields said:

Thank you for your reply @fresh_42 !

That is very helpful! Does this mean since the integral is negative, then the area is negative, so we can have negative area above the x-axis (without going below the x-axis)?

Many thanks!

It means that we need to be very careful when we consider the value of an integral as an area.

If ##F(x)## is the anti-derivative of ##f(x),## i.e. ##F(x)'=f(x)## or ##\int f(x)\,dx=F(x), ## then
$$
\int_a^b f(x)\,dx=F(b)-F(a)\text{ or } \int_a^a f(x)\,dx=\int_a^bf(x)\,dx +\int_b^a f(x)\,dx= 0
$$

The integral also makes a difference between above and below the ##x##-axis.

Example:
$$
\int_{-90°}^{90°} \sin(x)\,dx =\left[-\cos(x)\right]_{-90°}^{90°}=- \cos( 90°) -(- \cos (-90°) )=-0+(-0)=0
$$
But the area is not zero, it is
\begin{align*}
&\text{Area below the sine curve from }-90°\text{ to } +90°=\left|\int_{-90°}^{0°} \sin(x)\,dx\right|+\left|\int_{0°}^{90°} \sin(x)\,dx\right|\\
&=|-\cos(0°)+\cos(-90°)|+|-\cos(90°)+\cos(0°)|\\
&=|-1+0|+|-0+1|=|-1|+|1|=2
\end{align*}

Areas are always positive, integrals not necessarily. The sign of an integral depends on which quadrant of the ##(x,y)##-plane we are in, and whether our integration is clockwise or counterclockwise. To calculate an area, we have to split an integral into portions. Say ##a < c <b## and ##f(c)=0.## then the area below ##f(x)## is given as
$$
A=\left|\int_a^c f(x)\,dx \right|+\left| \int_c^b f(x)\,dx\right|
$$

There is just a difference between an area and an integral.