MHB Finding the exact length of the curve (II)

[shamieh]

Find the exact length of the curve

$0 \le x \le 1$

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

$$
\therefore y' = 9\sqrt{x}$$

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

$$= \int ^1_0 \sqrt{1 + 81x} \, dx$$

$$
= \int^1_0 1 + 9\sqrt{x} \, dx$$

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

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

Thus getting: $$1 + 6 = 7? $$

 

[chisigma]

Find the exact length of the curve

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

$$
\therefore y' = 9\sqrt{x}$$

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

$$= \int ^1_0 \sqrt{1 + 81x} \, dx$$

$$
= \int^1_0 1 + 9\sqrt{x} \, dx$$

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

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

Thus getting: $$1 + 6 = 7? $$

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 $$\ne$$ ?

$$\sqrt{1} = 1$$ , $$\sqrt{81} = 9 $$, $$\sqrt{x} = \sqrt{x}$$

 

[chisigma]

How is that $$\ne$$ ?

$$\sqrt{1} = 1$$ , $$\sqrt{81} = 9 $$, $$\sqrt{x} = \sqrt{x}$$

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 $$\ne$$ to the solution i proposed... $$\sqrt{x}$$ is just itself still $$\sqrt{x}$$ all we did was square x to make it $$\sqrt{x}$$ . How can you say that x square rooted isn't = to $$\sqrt{x}$$

What am I not seeing here?

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

 

[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 $\sqrt{a+b} = \sqrt a + \sqrt b$.