Question

y = 1 + 6x^(3/2) ; 0<= x <= 1

Find the Length

Answer 1

I know I'm supposed to use
\[L=∫\sqrt{1+(dy/dx)^2}dx\]

Answer 2

right so can you get the derivative first

Answer 3

dy/dx = 0 +9*x^(1/2)
(dy/dx)^2 = 81*x
this is what you should be able to get first

Answer 4

Right, that's what I got so it's:

\[\int\limits_{0}^{1} \sqrt{1+81x} dx\]
or
\[\int\limits_{0}^{1}(1+81x)^{1/2} dx\]
Where to I go from here?

Answer 5

so far no problem, the last part is to intergrate

Answer 6

\[2/243(1+81x)^{3/2}|^1_0\]

Answer 7

Yeah, for some reason I just completely forgot about u substitution. Thanks for your help.