what is the average value of 3x² + 4x on the interval 2 ≤ x ≤ 6?

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(x)$ on the interval $[a,b]$ is given by $\bar{y}=\frac{1}{b - a}\int_{a}^{b}f(x)dx$. Here, $a = 2$, $b = 6$, and $f(x)=3x^{2}+4x$. So, $\bar{y}=\frac{1}{6 - 2}\int_{2}^{6}(3x^{2}+4x)dx=\frac{1}{4}\int_{2}^{6}(3x^{2}+4x)dx$.

Step2: Integrate term - by - term

We know that $\int(3x^{2}+4x)dx=3\int x^{2}dx + 4\int xdx$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $3\times\frac{x^{3}}{3}+4\times\frac{x^{2}}{2}+C=x^{3}+2x^{2}+C$.

Step3: Evaluate the definite integral

$\frac{1}{4}\left[x^{3}+2x^{2}\right]_{2}^{6}=\frac{1}{4}\left((6^{3}+2\times6^{2})-(2^{3}+2\times2^{2})\right)$. First, calculate $6^{3}+2\times6^{2}=216 + 72=288$ and $2^{3}+2\times2^{2}=8 + 8 = 16$. Then, $\frac{1}{4}(288 - 16)=\frac{1}{4}\times272 = 68$.

Answer:

68