# Math Help - Imaginary Area? Huh?

1. ## Imaginary Area? Huh?

I don't understand the answer to the following definite integral. Can anyone help me explain why it is the way it is?

$\int_0^2{\frac{1}{^3\sqrt{t-1}}}dt$

I think the easiest way to see my point on this is to use u-substitution (though it isn't necessary): $u=t-1$

$\int_{-1}^1{u^{\frac{-1}{3}}}du$

$(\frac{3u^{\frac{2}{3}}}{2})|_{-1}^1$

$\frac{3(1)^{\frac{2}{3}}}{2} - \frac{3(-1)^{\frac{2}{3}}}{2}$

And I imagine going from the last step to the next one is my problem....

$\frac{3*1}{2} - \frac{3*1}{2} = 0$

But the answer cannot be 0! If you think about this being an area-under-the-curve problem, then the area is infinite because of the asymptote at the discontinuity within the limits of integration.

My suspicion is that $(-1)^{\frac{2}{3}}$ is somehow imaginary, but if it is I don't know why.

Can anyone explain this in reasonably understandable terms?

2. Originally Posted by pflo
I don't understand the answer to the following definite integral. Can anyone help me explain why it is the way it is?

$\int_0^2{\frac{1}{^3\sqrt{t-1}}}dt$

I think the easiest way to see my point on this is to use u-substitution (though it isn't necessary): $u=t-1$

$\int_{-1}^1{u^{\frac{-1}{3}}}du$

$(\frac{3u^{\frac{2}{3}}}{2})|_{-1}^1$

$\frac{3(1)^{\frac{2}{3}}}{2} - \frac{3(-1)^{\frac{2}{3}}}{2}$

And I imagine going from the last step to the next one is my problem....

$\frac{3*1}{2} - \frac{3*1}{2} = 0$

But the answer cannot be 0! If you think about this being an area-under-the-curve problem, then the area is infinite because of the asymptote at the discontinuity within the limits of integration.

My suspicion is that $(-1)^{\frac{2}{3}}$ is somehow imaginary, but if it is I don't know why.

Can anyone explain this in reasonably understandable terms?
You integrated through a singularity. The integrand is undefined at $t = 1$. The integral is improper and has to be treated special, i.e.

$\int_0^1 \frac{1}{\sqrt[3]{t-1}}\,dt + \int_1^2 \frac{1}{\sqrt[3]{t-1}}\,dt$. If either diverges the entire integral diverges.

3. Of course. I see the improper integral now. Thanks.

But this still leaves me with the same dilemma:
Originally Posted by Danny
$\int_0^1 \frac{1}{\sqrt[3]{t-1}}\,dt + \int_1^2 \frac{1}{\sqrt[3]{t-1}}\,dt$. If either diverges the entire integral diverges.
$\frac{3(t-1)^\frac{2}{3}}{2}|_0^1 + \frac{3(t-1)^\frac{2}{3}}{2}|_1^2$

$[\frac{3(0)^\frac{2}{3}}{2} - \frac{3(-1)^\frac{2}{3}}{2}] + [\frac{3(1)^\frac{2}{3}}{2} - \frac{3(0)^\frac{2}{3}}{2}]$

$[0 - \frac{3}{2}] + [\frac{3}{2} - 0]$

$\frac{-3}{2} + \frac{3}{2} = 0$

Neither of the integrals diverges, but looking at the graph of the integrand the area cannot possibly be 0.

4. Originally Posted by pflo
Of course. I see the improper integral now. Thanks.

But this still leaves me with the same dilemma:

$\frac{3(t-1)^\frac{2}{3}}{2}|_0^1 + \frac{3(t-1)^\frac{2}{3}}{2}|_1^2$

$[\frac{3(0)^\frac{2}{3}}{2} - \frac{3(-1)^\frac{2}{3}}{2}] + [\frac{3(1)^\frac{2}{3}}{2} - \frac{3(0)^\frac{2}{3}}{2}]$

$[0 - \frac{3}{2}] + [\frac{3}{2} - 0]$

$\frac{-3}{2} - \frac{3}{2} = 0$

Neither of the integrals diverges, but looking at the graph of the integrand the area cannot possibly be 0.
There's symmetry in the problem which accounts for the cancellation. It's easier to see if we shift to the origin.

$\int_{-1}^1 \frac{1}{\sqrt[3]{x}}\,dx$

Do to the symmetry in the origin this will equal zero. To find the area, you'll to compute

$- \int_{-1}^0 \frac{1}{\sqrt[3]{x}}\,dx + \int_{0}^1 \frac{1}{\sqrt[3]{x}}\,dx$

5. Originally Posted by Danny
There's symmetry in the problem which accounts for the cancellation. It's easier to see if we shift to the origin.
I agree the symmetry is easier to see if we shift to the origin, that's why I thought the u-substitution method was better to look at in this case. But I don't see how the symmetry could be causing the cancellation. I thought area above the x-axis was positive and area below the x-axis was negative. In this case, the area of both integrals is above the x-axis.

Am I wrong? Is area to the right of the y-axis positive and to the left negative? Or is it something else about the symmetry? How can I tell?

6. Originally Posted by pflo
I agree the symmetry is easier to see if we shift to the origin, that's why I thought the u-substitution method was better to look at in this case. But I don't see how the symmetry could be causing the cancellation. I thought area above the x-axis was positive and area below the x-axis was negative. In this case, the area of both integrals is above the x-axis.

Am I wrong? Is area to the right of the y-axis positive and to the left negative? Or is it something else about the symmetry? How can I tell?
Let's consider an easier problem. Calculate the area between $y = x$ and the $x$-axis on $[-1,1]$ . A quick calculation gives

$
\int_{-1}^1 x\,dx = 0
$

But the area can't be zero. So why did this happen? The function $y=x$ is symmetrical about the origin. I.e. if $f(x) = x$ then $f(-x) = -f(x)$. So $\int_{-1}^0 x\,dx = - \int_0^1 x\,dx$ and $\int_{-1}^0 x\,dx + \int_0^1 x\,dx = 0$.