$$ \begin{array} {|c||c|c|c|c|} \hline t \text{ (hours)} & 4 & 7 & 12 & 15 \\ \hline R(t) \text{ (liters/hour)} & 6.5 & 6.2 & 5.9 & 5.6 \\ \hline \end{array} $$

A tank contains 30 liters of oil at time \(t=4\) hours. Oil is being pumped into the tank at a rate \(R(t)\), where \(R(t)\) is measured in liters per hour, and \(t\) is measured in hours. Selected values of \(R(t)\) are given in the table above. Using a right Riemann sum with three subintervals and data from the table, what is the approximation of the number of liters of oil that are in the tank at time \(t=15\) hours?

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The amount of oil at \(t=15\) is equal to the amount of oil at \(t=4\) plus \( \int \limits_4^{15} R(t)   dt \). We are not given the equation of \(R(t)\), but we can estimate the value of the definite integral using a right Riemann sum.

The right Riemann sum can be visualized below:

The areas of the three regions:

$$ \sum^3_{i=1} \Delta x \cdot f(x_i) $$ $$ = (7-4)\cdot6.2 + (12-7)\cdot5.9+ (15-12)\cdot 5.6 $$ $$ = 64.9 $$

Putting it all together:

$$ 30 \text{ liters} + \int \limits_4^{15} R(t)   dt $$ $$ \approx 30 \text{ liters} + \sum^3_{i=1} \Delta x \cdot f(x_i) $$ $$ = 30 + 64.9 $$ $$ = \boxed{94.9} $$