x=sqrt(t) , y=3t-1 , 0

Arc length of a curve C described by the parametric equations x=f(t) and y=g(t),a<=t<=b , where f'(t) and g'(t) are continuous on [a,b] and C is traversed exactly once as t increases from a to b , then the length of the curve C is given by:
L=int_a^bsqrt((dx/dt)^2+(dy/dt)^2)dt
Given parametric equations are :x=sqrt(t),y=(3t-1) , 0<=t<=1
x=sqrt(t)
=>dx/dt=1/2(t)^(1/2-1)
dx/dt=1/2t^(-1/2)
dx/dt=1/(2sqrt(t))
y=3t-1
dy/dt=3
Now let's evaluate the length of the arc by using the stated formula,
L=int_0^1sqrt((1/(2sqrt(t)))^2+3^2)dt
L=int_0^1sqrt(1/(4t)+9)dt
L=int_0^1sqrt((1+36t)/(4t))dt
L=int_0^1sqrt(1+36t)/sqrt(4t)dt
L=int_0^1 1/2sqrt(1+36t)/sqrt(t)dt
Take the constant out,
L=1/2int_0^1sqrt(1+36t)/sqrt(t)dt
Let's first evaluate the indefinite integral:intsqrt(1+36t)/sqrt(t)dt
Use integral substitution:u=6sqrt(t)
du=6(1/2)t^(1/2-1)dt
du=3/sqrt(t)dt
intsqrt(1+36t)/sqrt(t)dt=int1/3sqrt(1+u^2)du
=1/3intsqrt(1+u^2)du
Use the following standard integral from the integration tables:
intsqrt(u^2+a^2)du=1/2(usqrt(u^2+a^2)+a^2ln|u+sqrt(a^2+u^2)|)+C
=1/3[1/2(usqrt(u^2+1)+ln|u+sqrt(1+u^2)|)]+C
=1/6(usqrt(1+u^2)+ln|u+sqrt(1+u^2)|)+C
Substitute back:u=6sqrt(t)
=1/6(6sqrt(t)sqrt(1+36t)+ln|6sqrt(t)+sqrt(1+36t)|)+C
L=1/2[1/6(6sqrt(t)sqrt(1+36t)+ln|6sqrt(t)+sqrt(1+36t)|)]_0^1
L=1/12[6sqrt(t)sqrt(1+36t)+ln|6sqrt(t)+sqrt(1+36t)|]_0^1
L=1/12{[6sqrt(37)+ln|6+sqrt(37)|]-[ln|1|]}
L=1/12[6sqrt(37)+ln|6+sqrt(37)|]
Using calculator,
L=1/12[36.4965751818+ln12.0827625303]
L=1/12[38.9883550344]
L=3.2490295862
L~~3.249
Arc length of the curve on the given interval ~~3.249
 

Please find my solution in the attached image.

The equation for arc length in parametric coordinates is:
L=int_a^b sqrt((dx/dt)^2+(dy/dt)^2) dt
Where in this case:
dx/dt=d/dt (t^(1/2))=1/2t^(-1/2)
dy/dt=d/dt(3t-1)=3
a=0, b=1
Therefore
 
L=int_0^1 sqrt((1/2t^(-1/2))^2+(3)^2) dt
L=int_0^1 sqrt((1/4t^(-1))+9) dt
Solve this integral numerically.
L~~3.249