Find the length of the curve $y(x) = \int_1^x\sqrt{t^3 - 1} \, dt$, $1 \leq x \leq 4$.
Not sure if I should be using first principle theorem (having some trouble with that) or if there is a simpler/more obvious method.
Find the length of the curve $y(x) = \int_1^x\sqrt{t^3 - 1} \, dt$, $1 \leq x \leq 4$.
Not sure if I should be using first principle theorem (having some trouble with that) or if there is a simpler/more obvious method.
Hint: Use the standard formula for arclength.
Soon you will need $\dfrac{dy}{dx}$. For this, use the Fundamental Theorem of Calculus.
The relevant part of the Fundamental Theorem says that under suitable conditions, if $w=\int_a^x f(t)\,dt$, then $\dfrac{dw}{dx}=f(x)$.
Use the fact that Andre noted above and then search for the following definite integral: $\int_1^4x^{3/2}dx$ which gives you the length. It is $62/5$.