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

Answer:

$c =2 \sqrt{3}$

Step-by-step explanation:

Using Mean Value Theorem for integrals

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

$f(x) = y = \dfrac{x^2}{4}, [0,6]$

Hence;

$\int^b_a \ f(x) \ dx = f(c) (b-a)= \int^6_0 \dfrac{x^2}{4}dx = f(c) (6-0)$

From above, using the power rule for integration

Then;

$\int x^n dx = \dfrac{x^{n+1}}{x+1}+C$

Thus;

$f(c)(6) = \Big[ \dfrac{x^3}{3(4)} \Big]^6_0$

$6f(c) =\Big[ \dfrac{x^3}{12} \Big]^6_0 = (\dfrac{6^3}{12}-0)$

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

$6f(c) = (18)$

$f(c) = 3$

From;

$f(x) = \dfrac{x^2}{4} \implies f(c) = \dfrac{c^2}{4}$

Hence,

$\dfrac{c^2}{4}= 3$

$c^2 =12$

$c = \sqrt{12}$

$c = \sqrt{4 \times 3}$

$c =2 \sqrt{3}$