Quick check of my work? (Arc Length)

**Bethanny** · November 29th 2010, 03:58 AM

G'day! I'm supposed to set up integrals and evaluate the length of the curve for three problems. I just wanted to make sure I've got the set up done right, since evaluating should be relatively easy.

So, the formula I'm using is:

$\int_a^b{\sqrt{1+(f'(x))^2}}dx$

a) for the curve $y=e^xcosx$ , in the interval $0\leq{x}\leq{\pi}$

I got: $\int_0^{\pi}{\sqrt{1+e^{2x}(1-sin2x)}}dx$

b) for the curve: $y=2x+sinx$ , in $0\leq{x}\leq{4\pi}$

I got: $\int_0^{4\pi}{\sqrt{cos^2x+4cosx+5}}dx$

c) for the curve: $y=6x+cosx$ , in $0\leq{x}\leq{5\pi}$

I got: $\int_0^{5\pi}{\sqrt{cos^2x+12cosx+37}}dx$

Thank you very much, mates! Lemme know if you need the step in between!

**Prove It** · November 29th 2010, 04:22 AM

a) is correct.

b) is correct.

c) should actually be $\displaystyle \int_0^{5\pi}{\sqrt{\sin^2{x} - 12\sin{x} + 37}\,dx}$ .

**Bethanny** · November 30th 2010, 03:44 AM

Thanks, mate!

I wrote the wrong trigonometric function. It's not $cosx$ , it's $sinx$ !! It should read: $6x + sinx$ !

How about now? Is my answer right?

**Prove It** · November 30th 2010, 04:47 AM

Originally Posted by Bethanny

Thanks, mate!

I wrote the wrong trigonometric function. It's not $cosx$ , it's $sinx$ !! It should read: $6x + sinx$ !

How about now? Is my answer right?

Yes, if the original function is $\displaystyle 6x + \sin{x}$ then your arc length integral is fine.

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