1. find the average value of the function f(x) = 3x^2 - 2x + 1 on the interval 1,4.

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(x)$ on the interval $[a,b]$ is given by $f_{avg}=\frac{1}{b - a}\int_{a}^{b}f(x)dx$. Here, $a = 1$, $b = 4$ and $f(x)=3x^{2}-2x + 1$. So, $f_{avg}=\frac{1}{4 - 1}\int_{1}^{4}(3x^{2}-2x + 1)dx=\frac{1}{3}\int_{1}^{4}(3x^{2}-2x + 1)dx$.

Step2: Integrate term - by - term

We know that $\int(3x^{2}-2x + 1)dx=3\int x^{2}dx-2\int xdx+\int 1dx$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $3\times\frac{x^{3}}{3}-2\times\frac{x^{2}}{2}+x=x^{3}-x^{2}+x$.

Step3: Evaluate the definite integral

$\frac{1}{3}\left[(x^{3}-x^{2}+x)\big|_{1}^{4}\right]=\frac{1}{3}[(4^{3}-4^{2}+4)-(1^{3}-1^{2}+1)]$. First, calculate $4^{3}-4^{2}+4=64 - 16+4=52$ and $1^{3}-1^{2}+1=1$. Then, $\frac{1}{3}(52 - 1)=\frac{51}{3}=17$.

Answer:

$17$