the function f is defined by f(x)=x² + 3 - 4sin(x²). what is the average value of f on the interval -1,8…

Explanation:

Step1: Find the derivative of (f(x))

Using the sum - rule and chain - rule, if (f(x)=x^{2}+3 - 4\sin(x^{2})), then (f^{\prime}(x)=2x-8x\cos(x^{2})).

Step2: Recall the average - value formula

The average value of a function (y = g(x)) on the interval ([a,b]) is given by (\frac{1}{b - a}\int_{a}^{b}g(x)dx). Here, (g(x)=f^{\prime}(x)), (a=-1), and (b = 8), so the average value of (f^{\prime}(x)) on ([-1,8]) is (\frac{1}{8-(-1)}\int_{-1}^{8}(2x - 8x\cos(x^{2}))dx).

Step3: Split the integral

(\frac{1}{9}\left(\int_{-1}^{8}2x dx-\int_{-1}^{8}8x\cos(x^{2})dx\right)). For (\int_{-1}^{8}2x dx), using the power - rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we have (\int_{-1}^{8}2x dx=x^{2}\big|{-1}^{8}=8^{2}-(-1)^{2}=64 - 1=63). For (\int{-1}^{8}8x\cos(x^{2})dx), use substitution. Let (u = x^{2}), then (du = 2x dx). When (x=-1), (u = 1); when (x = 8), (u = 64). So (\int_{-1}^{8}8x\cos(x^{2})dx=4\int_{1}^{64}\cos(u)du=4[\sin(u)]_{1}^{64}=4(\sin(64)-\sin(1))).

Step4: Calculate the average value

(\frac{1}{9}\left(63-4(\sin(64)-\sin(1))\right)). Using a calculator, (\sin(64)\approx0.900), (\sin(1)\approx0.841). (\frac{1}{9}\left(63-4(0.900 - 0.841)\right)=\frac{1}{9}(63-4\times0.059)=\frac{1}{9}(63 - 0.236)=\frac{62.764}{9}\approx6.974).

Answer:

(6.974)