Question

Find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval. y = x²/4, [0, 6] (Enter your answers as a comma-separated list.) f(x)

Answer 1

`c =2 √(3)`

Explanation:

Using Mean Value Theorem for integrals

`\int^b_a \ f(x) \ dx = f(c) (b-a)`

`f(x) = y = (x^2)/(4), [0,6]`

Hence;

`\int^b_a \ f(x) \ dx = f(c) (b-a)= \int^6_0 (x^2)/(4)dx = f(c) (6-0)`

From above, using the power rule for integration

Then;

`\int x^n dx = (x^(n+1))/(x+1)+C`

Thus;

`f(c)(6) = \Big[ (x^3)/(3(4)) \Big]^6_0`

`6f(c) =\Big[ (x^3)/(12) \Big]^6_0 = ((6^3)/(12)-0)`

`6f(c) = (18-0)`

`6f(c) = (18)`

`f(c) = 3`

From;

`f(x) = (x^2)/(4) \implies f(c) = (c^2)/(4)`

Hence,

`(c^2)/(4)= 3`

`c^2 =12`

`c = √(12)`

`c = √(4 * 3)`

`c =2 √(3)`