A tank contains 50 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?​

Respuesta :

The volume of oil inside the tank at [tex]t = 15\,h[/tex] is approximately 66.556 liters.

How to estimate volume capacity by Riemann sum

In this question we must estimate the volume by Riemann sum, which consist in sum of all areas of the trapezoids of equal width in the volume rate versus time graph. The right Riemann sum with three intervals for this case is described below:

[tex]Q = \Sigma \limits_{i=0}^{2} \left\{R_{i}+\frac{1}{2}\cdot [R_{i+1}-R_{i}] \right\}\cdot \Delta t[/tex]

[tex]Q = \left[6.5+\frac{1}{2}\cdot (6.2-6.5) \right]\cdot (3.667)+\left[6.2+\frac{1}{2}\cdot (5.9-6.2)\right]\cdot (3.667) +\left[5.9+\frac{1}{2}\cdot (5.6-5.9) \right]\cdot (3.667)[/tex]

[tex]Q \approx 66.556\,L[/tex]

The volume of oil inside the tank at [tex]t = 15\,h[/tex] is approximately 66.556 liters. [tex]\blacksquare[/tex]

Remark

The table of the volume rate versus time is missing, all missing values of the table are included below:

[tex]R(4) = 6.5\,\frac{L}{h}[/tex], [tex]R(7.667) = 6.2\,\frac{L}{h}[/tex], [tex]R(11.333) = 5.9\,\frac{L}{h}[/tex], [tex]R(15) = 5.6\,\frac{L}{h}[/tex]

To learn more on Riemann sum, we kindly invite to check this verified question: https://brainly.com/question/21847158