integrate I(t)=2t^1/3

Question

 t is time and initial capital stock Kis 25.find the capital acummulation during the time intervals [0,1] and [1,3]

Answer 1

For I(t) equal to 2t^1/3, build the integral by setting t as the base

and raising this base to the power of (1/3 + 1) or 4/3.

Next write t^4/3 and add the given coefficient of 2 to obtain 2t^4/3.

Then stick another coefficient of 3/4 between the 2 and the t

to gain 2(3/4)t^4/3 or 1.5t^4/3.

Now differentiate 1.5t^4/3 to confirm that d(1.5t^4/3)/dt goes to

(4/3) × (1.5) × t^{(4/3 − 3/3)} or (6/3)t^(1/3) or 2t^1/3.

∫_{(from 0 to 1)}2t^1/3dt evaluates to [1.5t^4/3_{(from 0 to 1)}] or

1.5(1)^4/3 − 1.5(0)^4/3 or 1.5.

∫_{(from 1 to 3)}2t^1/3dt evaluates to [1.5t^4/3_{(from 1 to 3)}] or

1.5(3)^4/3 − 1.5(1)^4/3 or 1.5×3×3^1/3 − 1.5×1×1^1/3;

a calculator will crush this last down to 4.5×1.44224957 − 1.5

or 4.990123065.