Hailie Blevins

2022-06-29

What is the arclength of $f(x)=\sqrt{(x-1)(2x+2)}-2x$ on $x\in [6,7]$?

tennispopj8

Beginner2022-06-30Added 20 answers

Explanation:

We have

$f(x)=\sqrt{(x-1)\ast 2\ast (x+1)}-2x$

$f(x)=\sqrt{2}\ast \sqrt{{x}^{2}-1}-2x$

So we get the integral

${\int}_{6}^{7}\sqrt{1+(\sqrt{2}\ast \frac{x}{\sqrt{{x}^{2}-1}}-2{)}^{2}}dx$

With a numerical method we find

$\approx 1.15037$

We have

$f(x)=\sqrt{(x-1)\ast 2\ast (x+1)}-2x$

$f(x)=\sqrt{2}\ast \sqrt{{x}^{2}-1}-2x$

So we get the integral

${\int}_{6}^{7}\sqrt{1+(\sqrt{2}\ast \frac{x}{\sqrt{{x}^{2}-1}}-2{)}^{2}}dx$

With a numerical method we find

$\approx 1.15037$

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