How do you find the lengths of the curve #y=(4/5)x^(5/4)# for #0<=x<=1#?

We Need the formula #s=int_a^bsqrt(1+(f'(x))^2)dx#.

given #f(x)=4/5*x^(5/4)#
using that #(x^n)'=nx^(n-1)# we get

#f'(x)=4/5*5/4*x^(5/4-1)#

#f'(x)=x^(1/4)# so we have to integrate

#int_0^1sqrt(1+sqrt(x))dx#
to calculate the indefinite integtral we Substitute

#t=sqrt(1+sqrt(x))#

so #x=(t^2-1)^2#

#dx=4t(t^2-1)dt#

and we have #4*int t^2(t^2-1)dt=4int(t^4-t^2dt#

this is #4(t^5/5-t^3/3)+C=4/15(1+sqrt(x))^(3/2)(3sqrt(x)-2)+C#
and we have

#[4/5sqrt(1+sqrt(x))^5-4/3sqrt(1+sqrt(x))^3]_0^1=8/15(1+sqrt(2))#

Set the total length of the curve #C# over the interval #a and b# as #L_c#

Let us consider a minuscule portion of the curve that is so short and greatly magnified that it looks as though it is straight.

Set its length as #deltas#

Using Pythagoras we have #(deltas)^2=(deltax)^2+(deltay)^2#

#lim_(deltas->0)(deltas)^2->(ds)^2= (dx)^2+(dy)^2#

Take the square root of both sides

#ds=sqrt((dx)^2+(sy)^2)#

Multiply by 1 and you do not change the value. However 1 comes in many forms.

#ds=1xxsqrt((dx)^2+(sy)^2) color(white)("dd")->color(white)("dd")ds=dx/dxxxsqrt((dx)^2+(sy)^2) #

#color(white)("ddddddddddddddddddddd")->color(white)("d")ds=dxxxsqrt((dx)^2/(dx)^2+(dy)^2/(dx)^2) #

#color(white)("ddddddddddddddddddddd")->color(white)("d")ds=dxxxsqrt(1+(dy)^2/(dx)^2) #

But #(dy)^2/(dx)^2 # may be written as # (f'(x))^2 #

#color(white)("ddddddddddddddddddddd")->color(white)("d")ds=dxxxsqrt(1+(f'(x))^2 ) #

#color(white)("ddddddddddddddddddddd")->color(white)("d")ds=color(white)("dd")sqrt(1+(f'(x))^2 ) xxdx #

Multiply #ds# by 1 and use the dot instead of #xx#

#color(white)("ddddddddddddddddddddd")->color(white)("d")1.ds=color(white)("dd")sqrt(1+(f'(x))^2 ) .dx #

Sum up all the #ds's# using integration

#color(white)("ddddddddddd")->color(white)("d")int_a^b color(white)(.)1.ds=L_c=color(white)("dd")int_a^bsqrt(1+(f'(x))^2 ) .dx #

As in Sonnhard's solution