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Suppose my function is f(x)=$x^2-1$. The absolute value of this function is $\sqrt{(x^2-1)^2}$. So why doesn't the area of this function between -2 and 2 equal $\int_{-2}^2\sqrt{(x^2-1)^2}$?

The antiderivative exists and equals $\frac{x(x^2-3)\sqrt{(x^2-1)^2}}{3(x^2-1)}$ but evaluating this at 2 and -2 doesn't give the correct answer.

Note: I'm not asking about how to actually find the answer here. I know to do that I'd find the zeros (-1 and 1) and then add up all the pieces. I'm trying to understand why the above does not reflect the actual area.

C Shreve
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2 Answers2

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The function that you have given as an antiderivative is not continuous. It is piecewise continuous on $(-\infty,-1)$, $(-1,1)$, and $(1,\infty)$, and on each of these intervals, its derivative is $\sqrt{(x^2-1)}$, so that is why it seems like a valid antiderivative. But your antiderivative has jumps at $-1$ and at $1$. Take a look at the graph of it to see the jumps. So it can't be used in the Fundamental Theorem of Calculus.

You can fix it by adding a step function. Let $$s(x)=\frac23\left(\frac{x-1}{|x-1|}+\frac{x+1}{|x+1|}\right)=\begin{cases}-\frac43&x<-1\\0&-1<x<1\\\frac43&x>1\end{cases}$$ This function has $0$ derivative everywhere except at its discontinuities. If you add this to your antiderivative, it's discontinuities become removable, and you'll have a new antiderivative for which you can use the Fundamental Theorem of Calculus.

Here is a picture of $y=\frac{x(x^2-3)\sqrt{(x^2-1)^2}}{3(x^2-1)}$:

enter image description here

Note that for $x$ slightly larger than $1$, then $x^2-1$ is positive, and $\sqrt{(x^2-1)^2}=x^2-1$. $$\begin{align} \frac{x(x^2-3)\sqrt{(x^2-1)^2}}{3(x^2-1)} &=\frac{x(x^2-3)(x^2-1)}{3(x^2-1)}\\ &=\frac{x(x^2-3)}{3}\\ &\approx-\frac{2}{3} \end{align}$$ but for $x$ slightly smaller than $1$, then $x^2-1$ is negative, and $\sqrt{(x^2-1)^2}=1-x^2$. $$\begin{align} \frac{x(x^2-3)\sqrt{(x^2-1)^2}}{3(x^2-1)} &=\frac{x(x^2-3)(1-x^2)}{3(x^2-1)}\\ &=-\frac{x(x^2-3)}{3}\\ &\approx\frac{2}{3} \end{align}$$

Here is a picture of $y=G(x)=\frac{x(x^2-3)\sqrt{(x^2-1)^2}}{3(x^2-1)}+s(x)$:

enter image description here

For this last function $G$, $$ \begin{align} G(2)-G(-2)&=2-(-2)\\ &=4 \end{align} $$ which should be what you were expecting for the area you originally set out to find.

2'5 9'2
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  • I have graphed it out many times and it is my understanding that this function is continuous. Continuous in the strict Epsilon-Delta definition of continuity. So it's a continuous function whose integral does not reflect its area. I'm really trying to understand why this is. – C Shreve Apr 14 '15 at 04:55
  • It's not continuous. I'll add pictures. – 2'5 9'2 Apr 14 '15 at 05:07
  • @CShreve OK, pictures added. – 2'5 9'2 Apr 14 '15 at 05:16
  • Alex - thanks so much. I was using Mathematica which actually did yield a graph that seemed continuous. Can I ask what graphing utility you are using? – C Shreve Apr 14 '15 at 05:25
  • GeoGebra. But graphing is not what led to me seeing the discontinuity. The denominator shows that the proposed antiderivative is discontinuous at $\pm1$, so I though about $1\pm\epsilon$ for $x$ and how the output would behave, and algebraically worked out the one-sided limits. I added that explanation between the pictures. I should have included it originally. – 2'5 9'2 Apr 14 '15 at 05:28
  • @alex.jordan has graphed the same function that I described above with a diffrent value of $c_1$. – tomi Apr 14 '15 at 13:04
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Your antiderivative function is wrong.

It is best to think of $|x^2-1|$ as three separate functions for three different domains.

$f_1(x)=x^2-1$ for $x<=-1$

$f_2(x)=1-x^2$ for $-1<=x<=1$

$f_3(x)=x^2-1$ for $x=>1$

This is a piecewise continuous function.

Integrating gives

$g_1(x)=\frac {x(x^2-3)} 3+c_1$ for $x<=-1$

$g_2(x)=\frac {x(3-x^2)} 3+c_2$ for $-1<=x<=1$

$g_3(x)=\frac {x(x^2-3)} 3+c_3$ for $x=>1$

We need this antiderivative function to also be piecewise continuous, so we choose appropriate values of $c_1$, $c_2$ and $c_3$.

Let $c_1=0$ for simplicity.

$g_1(-1)=\frac 2 3$

$g_2(-1)=-\frac 2 3 + c_2$, so $c_2=\frac 4 3$

$g_2(1)=2$

$g_3(1)=-\frac 2 3 + c_3$, so $c_3=\frac 8 3$

Putting it all together gives

$g_1(x)=\frac {x(x^2-3)} 3$ for $x<=-1$

$g_2(x)=\frac {x(3-x^2)} 3+\frac 2 3$ for $-1<=x<=1$

$g_3(x)=\frac {x(x^2-3)} 3+\frac 8 3$ for $x=>1$

That should work for you now.

tomi
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  • Why does one require that the anti-derivative be continuous when evaluating the definite integral? The value of the constants plays no role in the value of the integral. – Mark Viola Apr 13 '15 at 22:48
  • The constants do play a role, but usually the subtraction process means that the constant cancels itself out. May help to think about the original function as velocity. The antiderivative is displacement. To know how far you have travelled over a particular time interval you just need to know the difference between displacements, but you don't need to know where you are measuring displacement from (which would be the constant term). – tomi Apr 13 '15 at 22:54
  • That reasoning is based on some "physical" interpretation of the anti-derivative, not a mathematical one. In general , there is no mathematical reason for requiring continuity for finding the value of a definite integral. – Mark Viola Apr 13 '15 at 23:17
  • Take an arbitrary lower bound $x=a$. Take a value $b=>a$. We define the integral from $a$ to $b$ as the area under the graph between the lines $x=a$ and $x=b$. For a piecewise continuous curve, as $b$ increases, the area under the graph will also change continuously. The antiderivative function must therefore also be continuous. – tomi Apr 13 '15 at 23:23
  • Note: I am only insisting on a continuous antiderivative function because the original function is continuous. – tomi Apr 13 '15 at 23:27
  • I believe that I understand your goal now. You want to evaluate the integral as $g(2)-g(-2)$ and not worry about splitting the integral into 3 regions. That's fine. – Mark Viola Apr 13 '15 at 23:29