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

In summary: The value of an area depends on the coordinate system we are using, while the value of an integral is always positive.
  • #1
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I am interested whether we can get a negative area above the x-axis when the lower limit of integration is larger than the upper limit of integration
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,
1681195465513.png

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!
 
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  • #2
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.
 
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  • #3
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!
 
  • #4
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##.
 
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  • #5
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.
 
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1. What does negative area above the x-axis represent when integrating x^2?

The negative area above the x-axis when integrating x^2 represents a region where the function is below the x-axis. This means that the function has negative values in that interval.

2. Why is there negative area above the x-axis when integrating x^2?

This occurs because the function x^2 is symmetric about the y-axis, meaning that for every positive value above the x-axis, there is an equal negative value below the x-axis. When these negative values are integrated, they result in a negative area above the x-axis.

3. How is the negative area above the x-axis calculated when integrating x^2?

The negative area above the x-axis can be calculated by taking the integral of x^2 over the interval where the function is below the x-axis. This can be done by breaking the interval into smaller parts and using the fundamental theorem of calculus to integrate each part separately. The sum of these integrals will give the total negative area above the x-axis.

4. Is the negative area above the x-axis important in understanding the behavior of the function x^2?

Yes, the negative area above the x-axis can provide important information about the behavior of the function x^2. It can indicate the presence of roots or points of inflection, where the function changes from being above the x-axis to below it or vice versa. It can also help in determining the overall shape of the function.

5. Can the negative area above the x-axis be used to find the average value of the function x^2?

Yes, the negative area above the x-axis can be used to find the average value of the function x^2. This is done by taking the total negative area above the x-axis and dividing it by the length of the interval over which the function is below the x-axis. This will give the average value of the function in that interval.

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