what is the average (mean) value of $3t^{3}-t^{2}$ over the interval $-1leq tleq2$?\na $\frac{11}{4}$\nb…

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(t)$ over the interval $[a,b]$ is given by $\bar{y}=\frac{1}{b - a}\int_{a}^{b}f(t)dt$. Here, $a=-1$, $b = 2$ and $f(t)=3t^{3}-t^{2}$. So, $\bar{y}=\frac{1}{2-(-1)}\int_{-1}^{2}(3t^{3}-t^{2})dt=\frac{1}{3}\int_{-1}^{2}(3t^{3}-t^{2})dt$.

Step2: Integrate term - by - term

We know that $\int(3t^{3}-t^{2})dt=3\times\frac{t^{4}}{4}-\frac{t^{3}}{3}+C=\frac{3}{4}t^{4}-\frac{1}{3}t^{3}+C$.

Step3: Evaluate the definite integral

$\frac{1}{3}\left[\frac{3}{4}t^{4}-\frac{1}{3}t^{3}\right]_{-1}^{2}=\frac{1}{3}\left[\left(\frac{3}{4}(2)^{4}-\frac{1}{3}(2)^{3}\right)-\left(\frac{3}{4}(-1)^{4}-\frac{1}{3}(-1)^{3}\right)\right]$. First, calculate $\frac{3}{4}(2)^{4}-\frac{1}{3}(2)^{3}=\frac{3}{4}\times16-\frac{8}{3}=12-\frac{8}{3}=\frac{36 - 8}{3}=\frac{28}{3}$. Second, calculate $\frac{3}{4}(-1)^{4}-\frac{1}{3}(-1)^{3}=\frac{3}{4}+\frac{1}{3}=\frac{9 + 4}{12}=\frac{13}{12}$. Then, $\frac{1}{3}\left(\frac{28}{3}-\frac{13}{12}\right)=\frac{1}{3}\times\frac{112 - 13}{12}=\frac{1}{3}\times\frac{99}{12}=\frac{11}{4}$.

Answer:

A. $\frac{11}{4}$