Question

Use the mean value theorem to determine the area of f(x)= 2x^2+3 on the interval [0 , 2]

Answer 1

The area of given function is 5.67 unit²

Explanation:

Given function f(x) as :

f(x) = 2 x² + 3

The interval in which f(x) lies [ 0 , 2 ]

Let The area for the curve = A

Or , A =
`(1)/(b - a)\int_(a)^(b) f(x) dx`

Or, A =
`(1)/(2 - 0)\int_(0)^(2) (2x^(2)+3) dx`

Or, A =
`(1)/(2)(\int_(0)^(2) (2x^(2)) dx + \int_(0)^(2) 3 dx )`

or, A =
`(1)/(2) [2( (2^(3)-0^(3))/(3))] + (1)/(2)[3 (2-0)]`

or, A =
`(1)/(2)` (
`(16)/(3)` + 6 )

Or, A =
`(1)/(2)` ×
`(34)/(3)`

∴ A = 5.67 unit²

Hence The area of given function is 5.67 unit² Answer