what is the average (mean) value of $3t^{3}-t^{2}$ over the interval $-1leq tleq2$?

Explanation:

Step1: Recall the 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 $\int(3t^{3}-t^{2})dt=3\int t^{3}dt-\int t^{2}dt$. Using the power rule for integration $\int t^{n}dt=\frac{t^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $3\times\frac{t^{4}}{4}-\frac{t^{3}}{3}+C=\frac{3t^{4}}{4}-\frac{t^{3}}{3}+C$.

Step3: Evaluate the definite integral

$\frac{1}{3}\left[\frac{3t^{4}}{4}-\frac{t^{3}}{3}\right]_{-1}^{2}=\frac{1}{3}\left[\left(\frac{3\times2^{4}}{4}-\frac{2^{3}}{3}\right)-\left(\frac{3\times(-1)^{4}}{4}-\frac{(-1)^{3}}{3}\right)\right]$. First, calculate the values inside the brackets: For $t = 2$: $\frac{3\times16}{4}-\frac{8}{3}=12-\frac{8}{3}=\frac{36 - 8}{3}=\frac{28}{3}$. For $t=-1$: $\frac{3}{4}+\frac{1}{3}=\frac{9 + 4}{3}=\frac{13}{12}$. Then, $\frac{1}{3}\left(\frac{28}{3}-\frac{13}{12}\right)=\frac{1}{3}\left(\frac{112 - 13}{12}\right)=\frac{1}{3}\times\frac{99}{12}=\frac{11}{4}$.

Answer:

$\frac{11}{4}$