Question

Given the function y=2x^3-3x^2. Find the average value of the function over the interval (-1,3)

Answer 1

Answer:

Average value of the function over the interval is 3.

Step-by-step explanation:

Average value of the function y = 2x³ - 3x² over the interval (-1, 3) will be defined by the area under curve for the given interval divided by width of the interval.

A = $\int\limits^3_ {-1} \, (2x^{3}-3x^{2})dx$

   = $[\frac{2x^{4}}{4}-\frac{3x^{3}}{3}]^{3}_{(-1)}$

   = $[\frac{x^{4}}{2}]^{3}_{-1}-[x^{3}]^{3}_{-1}$

   = $[\frac{3^{4}-(-1)^{4} }{2}]-[3^{3}-(-1)^{3}]$

   = $[\frac{81-1}{2}]-[27+1]$

   = $40-28$

   = 12

Width of the interval = 3 - (-1) = 4

Average value = $\frac{12}{4}=3$

Therefore, average value of the function over the given interval is 3.