Financial math- force of interest

Smith makes deposits of 100$ at time 0, and X $ at time 3. the fund grows at a force of interest f= (t^2)/100, t>0. The amount of interest earned from time 3 to time 6 is X. Calculate X.

Any help would be really appreciated.

Have you considered the basic formulation? You can do it in stages if it is easier to keep track.

$\int_{0}^{3}\frac{t^{2}}{100} \; dt = \frac{9}{100} = 0.09$

$100 \cdot e^{0.09} = 109.417$

$\int_{3}^{6}\frac{t^{2}}{100} \; dt = \frac{63}{100} = 0.63$

$(109.417 + X) \cdot e^{0.63} = (109.417 + X) + X$

Solve for X.

Note: You may wish to eliminate some rounding. There really is no need to use the three decimal places for intermediate results. I get an error of about 0.3¢ due to the unnecessary rounding. The integrals are exact to two decimal places, so that creates no additional rounding problem.

Hmpf! Let's see you do the next one. Only one for free.