Thread: Quick check of my work? (Arc Length)

1. Quick check of my work? (Arc Length)

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!

2. a) is correct.

b) is correct.

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

3. Thanks, mate!

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

I wrote the wrong trigonometric function. It's not $cosx$, it's $sinx$!! It should read: $6x + sinx$!
Yes, if the original function is $\displaystyle 6x + \sin{x}$ then your arc length integral is fine.