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So i have a function: $$f(x)=2\sqrt{1-3x}$$

And i want to calculate the length of the curve in the first quadrant. So i thought of using it in these way: $$\int_{0}^{\frac{1}{3}}\sqrt{1+(f'(x))^2}$$

And i got till this derivative: $$f'(x)=2(1-3x)^{-\frac{1}{2}}$$

and so that integral looks like this:

$$\int_{0}^{\frac{1}{3}}\sqrt{1+(\frac{-6}{\sqrt{1-3x}})^2}dx$$ So i want to know whether i did this correctly, and i am having problems calculating this integral, i don't know how to approach it. Any help would be appreciated. Thank you in advance. And if i did any mistakes until now, correct me.

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  • $\begingroup$ type your formuals in proper $tex$ then no such problems appear $\endgroup$ – tired Aug 28 '16 at 10:39
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    $\begingroup$ well i can do that in the edit $\endgroup$ – MathIsTheWayOfLife Aug 28 '16 at 10:41
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    $\begingroup$ You don't have the correct expression for $f'(x)$. It should be $\frac{-3}{\sqrt{1-3x}}$ $\endgroup$ – David Quinn Aug 28 '16 at 11:19
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Let us make the problem more general, considering $$f(x)=a \sqrt{1-b x}\implies f'(x)=-\frac{a b}{2 \sqrt{1-b x}}$$ So $$I=\int \sqrt{1+\left(f'(x)\right)^2}\,dx=\int\sqrt{1+\frac{a^2 b^2}{4 (1-b x)}}\,dx$$ To get rid of the radical, let us set $$1+\frac{a^2 b^2}{4 (1-b x)}=t^2\implies x=\frac{-a^2 b^2+4 t^2-4}{4 b \left(t^2-1\right)}\implies dx=\frac{a^2 b t}{2 \left(1-t^2\right)^2}\,dt$$ which makes $$I=\frac{a^2 b} 2\int\frac{ t^2}{ \left(1-t^2\right)^2}\,dt$$ Now, as Jan Eerland answered, partial fraction decomposition.

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We get:

$$ \int_{0}^{1/3}\sqrt{1+(6(1-3x)^{-1/2})^2}dx= \int_{0}^{1/3}\sqrt{1+\dfrac{36}{1-3x}}dx= \int_{0}^{1/3}\sqrt{\dfrac{37-3x}{1-3x}}dx $$ $$ =\dfrac{1}{3}(\sqrt{37}+36 \ arccsch(6). $$

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  • $\begingroup$ The problem is that $6$ is wrong. $\endgroup$ – Claude Leibovici Aug 28 '16 at 14:30
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When $f(x)$ is the function, where you wnat to know the arc length from, use the formula:

$$\text{L}=\int_a^b\sqrt{1+\left(f'(x)\right)^2}\space\text{d}x$$


So, we get when $f(x)=2\sqrt{1-3x}$:

  1. $$f'(x)=\frac{\text{d}}{\text{d}x}\left(2\sqrt{1-3x}\right)=-\frac{3}{\sqrt{1-3x}}$$
  2. $$\left(f'(x)\right)^2=\left(-\frac{3}{\sqrt{1-3x}}\right)^2=\frac{9}{1-3x}$$
  3. $$\sqrt{1+\left(f'(x)\right)^2}=\sqrt{1+\frac{9}{1-3x}}=\sqrt{\frac{10-3x}{1-3x}}$$
  4. For the integral, use a substitution $u=-3x$ and $\text{d}u=-3\space\text{d}x$: $$\int\sqrt{1+\left(f'(x)\right)^2}\space\text{d}x=-\frac{1}{3}\int\sqrt{\frac{10+u}{1+u}}\space\text{d}u$$
  5. For the integral, use a substitution $s=\frac{10+u}{1+u}$ and $\text{d}s=\left(\frac{1}{1+u}-\frac{10+u}{(1+u)^2}\right)\space\text{d}u$: $$\int\sqrt{1+\left(f'(x)\right)^2}\space\text{d}x=-\frac{1}{3}\int\sqrt{\frac{10+u}{1+u}}\space\text{d}u=3\int\frac{\sqrt{s}}{(1-s^2)^2}\space\text{d}s$$
  6. For the integral, use a substitution $p=\sqrt{s}$ and $\text{d}p=\frac{1}{2\sqrt{s}}\space\text{d}s$: $$\int\sqrt{1+\left(f'(x)\right)^2}\space\text{d}x=-\frac{1}{3}\int\sqrt{\frac{10+u}{1+u}}\space\text{d}u=3\int\frac{\sqrt{s}}{(1-s^2)^2}\space\text{d}s=6\int\frac{p^2}{(1-p^2)^2}\space\text{d}p$$

Now, use partial fractions.

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  • $\begingroup$ what about the 2 before the square root in the derivative? it's as coefficient, not as constant right? $\endgroup$ – MathIsTheWayOfLife Aug 28 '16 at 10:47
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    $\begingroup$ @MathIsTheWayOfLife Notice that the derivative of $\sqrt{1-3x}$ equals $-\frac{3}{2\sqrt{1-3x}}$. Multiply that by 2 and you will get .... $\endgroup$ – Jan Eerland Aug 28 '16 at 11:24
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Since $y=2\sqrt{1-3x}$, solving for $x$ gives $y^2=4(1-3x)$ and $x=\frac{1}{3}(1-\frac{y^2}{4})$.

Then $\displaystyle s=\int_0^2\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy=\int_0^2\sqrt{1+\frac{y^2}{36}}dy=\frac{1}{6}\int_0^2\sqrt{y^2+36}dy$,

and letting $y=6\tan u, \;dy=6\sec^2udu$ gives

$\displaystyle6\int_{y=0}^{y=2}\sec^3 u\,du=6\left[\frac{1}{2}\big(\sec u\tan u+\ln|\sec u+\tan u|\big)\right]_{y=0}^{y=2}=3\left[\frac{y\sqrt{y^2+36}}{36}+\ln\left|\frac{\sqrt{y^2+36}+y}{6}\right|\right]_0^2=\color{blue}{\frac{\sqrt{10}}{3}+3\ln\left(\frac{\sqrt{10}+1}{3}\right)}$

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