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anonymous

  • one year ago

How do i find the arc length of: INT[1,x] sqrt(t^3-1) dt on 1<x<4?

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  1. freckles
    • one year ago
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    so you want to find the arclength of \[f(x)=\int\limits _1^x \sqrt{t^3-1} dt \text{ on } 1<x<4 \\ \text{ recall the arclength is given by } \\ L(x)=\int\limits_a^b \sqrt{1+(f'(x))^2} dx\]

  2. freckles
    • one year ago
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    can you find f'(x)?

  3. anonymous
    • one year ago
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    f'(x)=\[\frac{ 1 }{ 2 }x^2+\frac{ 1 }{ 2 }\]

  4. freckles
    • one year ago
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    \[f'(x)=\sqrt{x^3-1} \text{ by fundamental theorem of caluclus }\]

  5. freckles
    • one year ago
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    \[L=\int\limits_1^4 \sqrt{1+(\sqrt{x^3-1})^2} dx \\ \\ \text{ note : didn't mean to put } L(x) \text{ earlier } \\ L=\int\limits _1^4 \sqrt{1+(x^3-1)} dx\]

  6. freckles
    • one year ago
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    not entirely sure how you found your f'

  7. freckles
    • one year ago
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    @xnefop you there?

  8. anonymous
    • one year ago
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    Oh thank you i understand know. I forgot to apply the first part of the fundamental theorem.

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