Question

How do i find the arc length of: INT[1,x] sqrt(t^3-1) dt on 1 10 years ago

Answer 1

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

Answer 2

can you find f'(x)?

Answer 3

f'(x)=\[\frac{ 1 }{ 2 }x^2+\frac{ 1 }{ 2 }\]

Answer 4

\[f'(x)=\sqrt{x^3-1} \text{ by fundamental theorem of caluclus }\]

Answer 5

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

Answer 6

not entirely sure how you found your f'

Answer 7

@xnefop you there?

Answer 8

Oh thank you i understand know. I forgot to apply the first part of the fundamental theorem.