# Finding the exact length of the curve (II)

#### shamieh

##### Active member
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: $$\displaystyle 1 + 6 = 7?$$

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#### chisigma

##### Well-known member
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

##### Active member
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

##### Well-known member
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

##### Active member
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

##### Well-known member
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

##### MHB Oldtimer
Staff member
... in other words, it is NOT TRUE that $\sqrt{a+b} = \sqrt a + \sqrt b$.