integral question

**Sneaky** · September 16th 2011, 10:05 PM

How do I start this integral?
integral from [-2 to 1]
sqrt((x^2)-1) / x dx

**Prove It** · September 16th 2011, 10:33 PM

Originally Posted by Sneaky

How do I start this integral?
integral from [-2 to 1]
sqrt((x^2)-1) / x dx

$\displaystyle \int_{-2}^1{\frac{\sqrt{x^2 - 1}}{x}\,dx} = \frac{1}{2}\int_{-2}^1{\frac{2x\sqrt{x^2 - 1}}{x^2}\,dx}$

Now make the substitution $\displaystyle u = x^2 - 1 \implies du = 2x\,dx$ , and making note that $\displaystyle u(-2) = 3, u(1) = 0$ , the integral becomes

$\displaystyle \begin{align*} \frac{1}{2}\int_{-2}^1{\frac{2x\,\sqrt{x^2 - 1}}{x^2}\,dx} &= \frac{1}{2}\int_3^0{\frac{\sqrt{u}}{u + 1}\,du} \\ &= -\frac{1}{2}\int_0^3{\frac{\sqrt{u}}{u+1}\,du} \\ &= -\int_0^3{\frac{u}{2\sqrt{u}\,(u + 1)}\,du} \end{align*}$

Now make the substitution $\displaystyle v = \sqrt{u} \implies dv = \frac{1}{2\sqrt{u}}\,du$ and noting that $\displaystyle v(0) = 0$ and $\displaystyle v(3) = \sqrt{3}$ , the integral becomes

$\displaystyle \begin{align*} -\int_0^3{\frac{u}{2\sqrt{u}\,(u + 1)}\,du} &= -\int_0^{\sqrt{3}}{\frac{v^2}{v^2 + 1}\,dv} \\ &= -\int_0^{\sqrt{3}}{1 - \frac{1}{v^2 + 1}\,dv} \\ &= -\left[v - \arctan{v}\right]_0^{\sqrt{3}} \\ &= -\left[\left(\sqrt{3} - \arctan{\sqrt{3}}\right) - \left(0 - \arctan{0}\right)\right] \\ &= -\left(\sqrt{3} - \frac{\pi}{3} - 0 + 0\right) \\ &= \frac{\pi}{3} - \sqrt{3} \end{align*}$

**skeeter** · September 17th 2011, 05:39 AM

Originally Posted by Sneaky

How do I start this integral?
integral from [-2 to 1]
sqrt((x^2)-1) / x dx

please recheck the given limits of integration. the integrand, $\frac{\sqrt{x^2-1}}{x}$ , is undefined on the interval $-1 < x < 1$

**Sneaky** · September 17th 2011, 10:02 AM

oh i mis-read it, its -2 to -1. But I think I can fix it myself.
Also I have a question about how to change the integral bounds when you do a change of variable. What if you change u=cos(x), and the integral bound was -inf to inf, then what would be the new integral bounds?

**Prove It** · September 17th 2011, 10:09 AM

Originally Posted by Sneaky

oh i mis-read it, its -2 to -1. But I think I can fix it myself.
Also I have a question about how to change the integral bounds when you do a change of variable. What if you change u=cos(x), and the integral bound was -inf to inf, then what would be the new integral bounds?

I expect you would not ever have such an integral...

**Sneaky** · September 17th 2011, 10:23 AM

For the integral ln(x) / x dx, i let u=ln(x), then got
[e to infinity]

integral u du
[1 to infinity]

then i got (after integrating and subbing in the bounds)
ln^2(infinity)/2 - 0

Is this right so far?

**Prove It** · September 17th 2011, 10:29 AM

If you change the bounds, you don't convert the function back to a function of x, leave it in terms of u and substitute the u values.

But you are correct that the integral does not converge.

**Sneaky** · September 17th 2011, 10:32 AM

If i do that I end up with (u^2) / 2 |[1 to inf]
then it ends up being infinity - 1/2
It doesn't look correct.

**Sneaky** · September 17th 2011, 04:21 PM

Does this mean the answer is infinity?

**Prove It** · September 17th 2011, 10:02 PM

Originally Posted by Sneaky

Does this mean the answer is infinity?

It means that the integral does not converge... Infinity is not a number...

Thread: integral question

LinkBack

Thread Tools

Display