3. the rate at which rain accumulates in a bucket is modeled by the function r given by ( r(t)=10 t - t^{2}…

3. the rate at which rain accumulates in a bucket is modeled by the function r given by ( r(t)=10 t - t^{2} ), where ( r(t) ) is measured in milliliters per minute and ( t ) is measured in minutes since the rain began falling. how many milliliters of rain accumulate in the bucket from time ( t = 0 ) to time ( t = 3 )? (a) 4 (b) 21 (c) 36 (d) 63

3. the rate at which rain accumulates in a bucket is modeled by the function r given by ( r(t)=10 t - t^{2} ), where ( r(t) ) is measured in milliliters per minute and ( t ) is measured in minutes since the rain began falling. how many milliliters of rain accumulate in the bucket from time ( t = 0 ) to time ( t = 3 )? (a) 4 (b) 21 (c) 36 (d) 63

Answer

Explanation:

Step1: Recall the definite - integral formula for accumulation

The amount of rain accumulated from (t = a) to (t = b) is given by (A=\int_{a}^{b}r(t)dt). Here, (a = 0), (b = 3), and (r(t)=10t - t^{2}). So, (A=\int_{0}^{3}(10t - t^{2})dt).

Step2: Integrate term - by - term

Use the power rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)). (\int(10t - t^{2})dt=10\int tdt-\int t^{2}dt). For (\int tdt=\frac{t^{2}}{2}) and (\int t^{2}dt=\frac{t^{3}}{3}). So, (10\int tdt-\int t^{2}dt = 10\times\frac{t^{2}}{2}-\frac{t^{3}}{3}=5t^{2}-\frac{t^{3}}{3}+C).

Step3: Evaluate the definite integral

By the fundamental theorem of calculus (\int_{0}^{3}(10t - t^{2})dt=\left[5t^{2}-\frac{t^{3}}{3}\right]{0}^{3}). Substitute (t = 3) and (t = 0): When (t = 3), (5\times3^{2}-\frac{3^{3}}{3}=5\times9 - 9=45 - 9 = 36). When (t = 0), (5\times0^{2}-\frac{0^{3}}{3}=0). Then (\left[5t^{2}-\frac{t^{3}}{3}\right]{0}^{3}=36-0 = 36).

Answer:

C. 36