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.
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.
$\displaystyle \int_{0}^{3}\frac{t^{2}}{100} \; dt = \frac{9}{100} = 0.09$
$\displaystyle 100 \cdot e^{0.09} = 109.417$
$\displaystyle \int_{3}^{6}\frac{t^{2}}{100} \; dt = \frac{63}{100} = 0.63$
$\displaystyle (109.417 + X) \cdot e^{0.63} = (109.417 + X) + X$
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.