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

Question

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

freckles · Answer

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

freckles · Answer

can you find f'(x)?

anonymous · Answer

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

freckles · Answer

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

freckles · Answer

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

freckles · Answer

not entirely sure how you found your f'

freckles · Answer

@xnefop you there?

anonymous · Answer

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