Question

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

Answer 1

Average value of the function over the interval is 3.

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`

=
`[(2x^(4))/(4)-(3x^(3))/(3)]^(3)_((-1))`

=
`[(x^(4))/(2)]^(3)_(-1)-[x^(3)]^(3)_(-1)`

=
`[(3^(4)-(-1)^(4) )/(2)]-[3^(3)-(-1)^(3)]`

=
`[(81-1)/(2)]-[27+1]`

=
`40-28`

= 12

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

Average value =
`(12)/(4)=3`

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