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

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:

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