find the average value of y over the interval -1, 2.\ny = .1x^3 + 1\navg = ?\nround your answer to the…

Explanation:

Step1: State the formula for the average value of a function.

The average value of a function $f(x)$ over the interval $[a, b]$ is given by the formula: $$ AVG = \frac{1}{b-a} \int_{a}^{b} f(x) dx $$

Step2: Identify the function and the interval.

The given function is $y = f(x) = 0.1x^3 + 1$. The interval is $[a, b] = [-1, 2]$.

Step3: Set up the definite integral for the average value.

Substitute the function and interval limits into the formula: $$ AVG = \frac{1}{2 - (-1)} \int_{-1}^{2} (0.1x^3 + 1) dx $$ $$ AVG = \frac{1}{3} \int_{-1}^{2} (0.1x^3 + 1) dx $$

Step4: Evaluate the definite integral.

First, find the antiderivative of $f(x) = 0.1x^3 + 1$: $$ \int (0.1x^3 + 1) dx = 0.1 \frac{x^4}{4} + x = 0.025x^4 + x $$ Now, apply the Fundamental Theorem of Calculus: $$ \int_{-1}^{2} (0.1x^3 + 1) dx = [0.025x^4 + x]_{-1}^{2} $$ $$ = (0.025(2)^4 + 2) - (0.025(-1)^4 + (-1)) $$ $$ = (0.025(16) + 2) - (0.025(1) - 1) $$ $$ = (0.4 + 2) - (0.025 - 1) $$ $$ = 2.4 - (-0.975) $$ $$ = 2.4 + 0.975 = 3.375 $$

Step5: Calculate the average value.

Substitute the value of the integral back into the average value formula: $$ AVG = \frac{1}{3} (3.375) $$ $$ AVG = \frac{3.375}{3} = 1.125 $$

Step6: Round the answer to the nearest thousandth.

The calculated average value is $1.125$. This value is already expressed to the thousandth place.

Answer:

$1.125$