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Andreyy89
4 years ago
12

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

Mathematics
1 answer:
MArishka [77]4 years ago
8 0

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.

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