c1 - average value of a function: problem 1\n(1 point)\nthe average value of ( f(x)=3 x^{3}-21 x^{2} ) on…

Explanation:

Step1: Recall the formula for the average value of a function

The formula for the average value of a function (y = f(x)) on the interval ([a,b]) is (f_{avg}=\frac{1}{b - a}\int_{a}^{b}f(x)dx). Here, (a=-4), (b = 7), and (f(x)=3x^{3}-21x^{2}). So, (f_{avg}=\frac{1}{7-(-4)}\int_{-4}^{7}(3x^{3}-21x^{2})dx=\frac{1}{11}\int_{-4}^{7}(3x^{3}-21x^{2})dx).

Step2: Integrate term - by - term

We know that (\int(3x^{3}-21x^{2})dx=3\int x^{3}dx-21\int x^{2}dx). Using the power rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we have (3\times\frac{x^{4}}{4}-21\times\frac{x^{3}}{3}=\frac{3}{4}x^{4}-7x^{3}+C).

Step3: Evaluate the definite integral

(\frac{1}{11}\left[\frac{3}{4}x^{4}-7x^{3}\right]_{-4}^{7}=\frac{1}{11}\left[\left(\frac{3}{4}(7)^{4}-7(7)^{3}\right)-\left(\frac{3}{4}(-4)^{4}-7(-4)^{3}\right)\right]). First, calculate (\frac{3}{4}(7)^{4}-7(7)^{3}): (\frac{3}{4}(7)^{4}-7(7)^{3}=7^{3}\left(\frac{3\times7}{4}-7\right)=343\left(\frac{21 - 28}{4}\right)=343\times\left(-\frac{7}{4}\right)=-\frac{2401}{4}). Second, calculate (\frac{3}{4}(-4)^{4}-7(-4)^{3}): (\frac{3}{4}(-4)^{4}-7(-4)^{3}=\frac{3}{4}\times256-7\times(-64)=192 + 448=640). Then (\frac{1}{11}\left(-\frac{2401}{4}-640\right)=\frac{1}{11}\left(-\frac{2401+2560}{4}\right)=\frac{1}{11}\left(-\frac{4961}{4}\right)=-\frac{451}{4}=- 112.75).

Answer:

(-112.75)