this question is designed to be answered without a calculator. the rate, in liters per minute, at which…

Explanation:

Step1: Recall the definite - integral formula

The total amount of water pumped $A$ from $t = a$ to $t = b$ is given by $A=\int_{a}^{b}r(t)dt$. Here, $r(t)=t^{2}+1$, $a = 0$, and $b = 3$. So, $A=\int_{0}^{3}(t^{2}+1)dt$.

Step2: Use the integral sum rule

By the sum - rule of integration $\int_{a}^{b}(f(t)+g(t))dt=\int_{a}^{b}f(t)dt+\int_{a}^{b}g(t)dt$. So, $\int_{0}^{3}(t^{2}+1)dt=\int_{0}^{3}t^{2}dt+\int_{0}^{3}1dt$.

Step3: Apply the power - rule of integration

The power - rule for integration is $\int t^{n}dt=\frac{t^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $\int_{0}^{3}t^{2}dt$, we have $\left[\frac{t^{3}}{3}\right]{0}^{3}=\frac{3^{3}}{3}-\frac{0^{3}}{3}=9$. For $\int{0}^{3}1dt$, we have $[t]_{0}^{3}=3 - 0=3$.

Step4: Calculate the total integral

$\int_{0}^{3}(t^{2}+1)dt=9 + 3=12$.

Answer:

12 liters.