Finding the exact length of the curve (II)

[shamieh]

Find the exact length of the curve

$0 \le x \le 1$

\(\displaystyle y = 1 + 6x^{\frac{3}{2}}\) <-- If you can't read this, the exponent is \(\displaystyle \frac{3}{2}\)

\(\displaystyle
\therefore y' = 9\sqrt{x}\)

\(\displaystyle \int ^1_0 \sqrt{1 + (9\sqrt{x})^2} \, dx\)

\(\displaystyle = \int ^1_0 \sqrt{1 + 81x} \, dx\)

\(\displaystyle
= \int^1_0 1 + 9\sqrt{x} \, dx\)

Now can't I just split the two integrals separately to obtain:

\(\displaystyle x + 6x^{\frac{3}{2}} |^1_0 \) <-- If you can't read this, the exponent is \(\displaystyle \frac{3}{2}\)

Thus getting: \(\displaystyle 1 + 6 = 7? \)

 

[chisigma]

Find the exact length of the curve

\(\displaystyle y = 1 + 6x^{\frac{3}{2}}\) <-- If you can't read this, the exponent is \(\displaystyle \frac{3}{2}\)

\(\displaystyle
\therefore y' = 9\sqrt{x}\)

\(\displaystyle \int ^1_0 \sqrt{1 + (9\sqrt{x})^2} \, dx\)

\(\displaystyle = \int ^1_0 \sqrt{1 + 81x} \, dx\)

\(\displaystyle
= \int^1_0 1 + 9\sqrt{x} \, dx\)

Now can't I just split the two integrals separately to obtain:

\(\displaystyle x + 6x^{\frac{3}{2}} |^1_0 \) <-- If you can't read this, the exponent is \(\displaystyle \frac{3}{2}\)

Thus getting: \(\displaystyle 1 + 6 = 7? \)

There is a little questionable point because $\displaystyle \sqrt{1 + 81\ x} \ne 1 + 9\sqrt{x}$...

Kind regards

$\chi$ $\sigma$

 

[shamieh]

There is a little questionable point because $\displaystyle \sqrt{1 + 81\ x} \ne 1 + 9\sqrt{x}$...

Kind regards

$\chi$ $\sigma$

How is that \(\displaystyle \ne\) ?

\(\displaystyle \sqrt{1} = 1\) , \(\displaystyle \sqrt{81} = 9 \), \(\displaystyle \sqrt{x} = \sqrt{x}\)

 

[chisigma]

How is that \(\displaystyle \ne\) ?

\(\displaystyle \sqrt{1} = 1\) , \(\displaystyle \sqrt{81} = 9 \), \(\displaystyle \sqrt{x} = \sqrt{x}\)

The symbol $\ne$ means 'not equal to'...

Kind regards

$\chi$ $\sigma$

 

[shamieh]

The symbol $\ne$ means 'not equal to'...

Kind regards

$\chi$ $\sigma$

I understand what the symbol means, my question is tho, why is it \(\displaystyle \ne\) to the solution i proposed... \(\displaystyle \sqrt{x}\) is just itself still \(\displaystyle \sqrt{x}\) all we did was square x to make it \(\displaystyle \sqrt{x}\) . How can you say that x square rooted isn't = to \(\displaystyle \sqrt{x}\)

What am I not seeing here?

because if we have x then decide to square root x , we will just end up with \(\displaystyle \sqrt{x}\)

 

[chisigma]

What I meant to say is that You wrote something like...

$\displaystyle \int_{0}^{1} \sqrt{1 + 81\ x}\ dx = \int_{0}^{1} (1 + 9\ \sqrt{x})\ dx\ (1)$

... but it isn't true because [as You can easily verify...] for $0 < x \le 1$ is...

$\displaystyle \sqrt{1 + 81\ x} \ne 1 + 9\ \sqrt{x}\ (2)$

Kind regards

$\chi$ $\sigma$

 

[Opalg]

... in other words, it is NOT TRUE that $\sqrt{a+b} = \sqrt a + \sqrt b$.