Taking the Integral - did I do it right?

**funnytim** · September 18th 2010, 02:26 AM

Hey everyone,

I'm needing to take the integral of the following. However I don't get the correct answer (when I put in the bounds), so I'm wondering if I integrated it wrong somewhere?

The equation is
$\int_2^3 \sqrt{1 + (x^2/8 -2/x^2)^2}$

So I started off by multiplying out the square,
$\int_2^3 \sqrt{1 + x^4/64 - 2x^2/8x^2 - 2 x^2/8x^2 + 4/x^4}$

=
$\int_2^3 \sqrt{1 + x^4/64 - 1/2 + 4/x^4}$

Square rooted everything:
$\int_2^3 {1 + x^3/8 - 1/\sqrt{2} + 2x^-3}$

Integrated it: $x +x^4/32 - x/ \sqrt{2} - x^-2$

Did I do something wrong up there?

Thanks in advance!

**Educated** · September 18th 2010, 02:53 AM

Step 4. You can't just take the square root of everything inside the square root. It doesn't work.

$\sqrt{4+9} \ne \sqrt{4} + \sqrt{9}$

You will need to use integrating rules and finding the antiderivative to solve it.

**Prove It** · September 18th 2010, 02:57 AM

You can't take "square roots" term by term.

I would do it this way...

$1 + \left(\frac{x^2}{8} - \frac{2}{x^2}\right)^2 = 1 + \left(\frac{x^4}{8x^2} - \frac{16}{8x^2}\right)$

$= 1 + \left(\frac{x^4 - 16}{8x^2}\right)^2$

$= 1 + \frac{x^8 - 32x^4 + 256}{64x^4}$

$= \frac{64x^4}{64x^4} + \frac{x^8 - 32x^4 + 256}{64x^4}$

$= \frac{x^8 + 32x^4 + 256}{64x^4}$

$= \frac{(x^4 + 16)^2}{(8x^2)^2}$

$= \left(\frac{x^4 + 16}{8x^2}\right)^2$ .

Therefore $\int{\sqrt{1 + \left(\frac{x^2}{8} - \frac{2}{x^2}\right)^2\,dx} = \int{\sqrt{\left(\frac{x^4 + 16}{8x^2}\right)^2}\,dx}$

$= \int{\frac{x^4 + 16}{8x^2}\,dx}$

$= \int{\frac{x^2}{8} + \frac{2}{x^2}\,dx}$

$= \int{\frac{1}{8}x^2 + 2x^{-2}\,dx}$

$= \frac{1}{8}\cdot \frac{x^3}{3} + 2\frac{x^{-1}}{-1} + C$

$= \frac{x^3}{24} - \frac{2}{x} + C$ .

Now substitute your terminals.

**Archie Meade** · September 18th 2010, 02:57 AM

Originally Posted by funnytim

Hey everyone,

I'm needing to take the integral of the following. However I don't get the correct answer (when I put in the bounds), so I'm wondering if I integrated it wrong somewhere?

The equation is
$\int_2^3 \sqrt{1 + (x^2/8 -2/x^2)^2}\ dx$

So I started off by multiplying out the square,
$\int_2^3 \left(\sqrt{1 + x^4/64 - 2x^2/8x^2 - 2 x^2/8x^2 + 4/x^4}\right)\ dx$

=
$\int_2^3 \left(\sqrt{1 + x^4/64 - 1/2 + 4/x^4}\right)\ dx$

Square rooted everything:
$\int_2^3 \left({1 + x^3/8 - 1/\sqrt{2} + 2x^{-3}}\right)\ dx$ this is incorrect

Integrated it: $x +x^4/32 - x/ \sqrt{2} - x^{-2}$

Did I do something wrong up there?

Thanks in advance!

$\displaystyle\sqrt{\frac{x^4}{64}+\frac{1}{2}+\fra c{4}{x^4}}=\sqrt{\left(\frac{x^2}{8}+\frac{2}{x^2} \right)^2}$

**funnytim** · September 22nd 2010, 12:20 AM

Right, it was the square root thingy that was wrong.

Thanks so much for your help guys

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