x=6t^2 , y=2t^3 , 1

The formula of arc length of a parametric equation on the interval alt=tlt=b is:
L = int_a^b sqrt((dx/dt)^2+(dy/dt)^2) dt
The given parametric equation is:
 x=6t^2
y=2t^3
The derivative of x and y are:
dx/dt= 12t
dy/dt = 6t^2
 So the integral needed to compute the arc length of the given parametric equation on the interval 1lt=tlt=4 is:
L= int_1^4 sqrt ((12t)^2+(6t^2)^2) dt
The simplified form of the integral is:
L= int_1^4 sqrt (144t^2+36t^4)dt
L=int_1^4 sqrt (36t^2(4+t^2))dt
L= int_1^4 6tsqrt(4+t^2)dt
To take the integral, apply u-substitution method.

u= 4+t^2
du=2t dt
1/2du=tdt
t=1 ,  u =4+1^2=5
t=4 ,  u = 4+4^2=20

Expressing the integral in terms of u, it becomes:
L=6int_1^4 sqrt(4+t^2)* tdt
L=6 int _5^20 sqrtu *1/2du
L=3int_5^20 sqrtu du
L=3int_5^20 u^(1/2)du
L=3*u^(3/2)/(3/2)  |_5^20
L=2u^(3/2)  |_5^20
L = 2usqrtu  |_5^20
L=2(20)sqrt20 - 2(5)sqrt5
L=40sqrt20-10sqrt5
L=40*2sqrt5 - 10sqrt5
L=80sqrt5-10sqrt5
L=70sqrt5
Therefore, the arc length of the curve is  70sqrt5  units.