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Math Help - Quick check of my work? (Arc Length)

  1. #1
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    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!
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  2. #2
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    a) is correct.

    b) is correct.

    c) should actually be \displaystyle \int_0^{5\pi}{\sqrt{\sin^2{x} - 12\sin{x} + 37}\,dx}.
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  3. #3
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    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?
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  4. #4
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    Quote Originally Posted by Bethanny View Post
    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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