How do you find the point c in the interval #0

Question

How do you find the point c in the interval #0<=x<=3# such that f(c) is equation to the average value of #f(x)=3x^2#?

HIX Tutor · Accepted Answer

undefined Alright! First, let's find the average value of f(x)=3x^2 on the interval [0,3].

1. Calculate the definite integral of f(x)=3x^2 on the interval [0,3]: 
∫[0 to 3] 3x^2 dx = [x^3] from 0 to 3 = 3^3 - 0^3 = 27

2. Find the length of the interval [0,3]: 
Length of interval = 3 - 0 = 3

3. Calculate the average value:
Average value = (1/(3-0)) * ∫[0 to 3] 3x^2 dx = 27/3 = 9

4. To find the point c on the interval [0,3] where f(c) is equal to the average value (9), set up the equation:
3c^2 = 9

5. Solve for c:
c^2 = 3
c = √3

Therefore, the point c in the interval 0<=x<=3 such that f(c) is equal to the average value of f(x)=3x^2 is c=√3.

Ezekiel Alexander · Answer

Find the average value. Set #f(x)# equal to the average value. Solve that equation in the interval #[0,3]#.

#f_"ave" = 1/(3-0) int_0^3 3x^2 dx# # = 1/3 [x^3]_0^3# # 1/3[(3^3)-(0^3)] = 1/3(27) = 9# Now solve #3x^2 = 9# #" "# in #[0,3]# #x^2=3# #x=+-sqrt3# #" "# Note that the algebra gives us two solutions, but one of them is not in the interval. #c = sqrt3#

Samantha Adkison · Answer

undefined To find the point $ c $ in the interval $ 0 \leq x \leq 3 $ such that $ f(c) $ is equal to the average value of $ f(x) = 3x^2 $, you first need to find the average value of $ f(x) $ over the interval $ [0,3] $, which is given by the formula:

$$ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx $$

Substitute $ a = 0 $ and $ b = 3 $ into the formula:

$$ \text{Average value} = \frac{1}{3-0} \int_{0}^{3} 3x^2 \, dx $$

$$ = \frac{1}{3} \int_{0}^{3} 3x^2 \, dx $$

$$ = \frac{1}{3} [x^3]_{0}^{3} $$

$$ = \frac{1}{3} (3^3 - 0^3) $$

$$ = \frac{1}{3} (27) $$

$$ = 9 $$

So, the average value of $ f(x) $ over the interval $ [0,3] $ is 9. Now, to find the point $ c $ where $ f(c) $ is equal to 9, you set $ f(c) = 3c^2 $ equal to 9 and solve for $ c $:

$$ 3c^2 = 9 $$

$$ c^2 = 3 $$

$$ c = \pm \sqrt{3} $$

Since $ c $ must be in the interval $ 0 \leq c \leq 3 $, the value of $ c $ is $ c = \sqrt{3} $.

How do you find the point c in the interval #0&lt;=x&lt;=3# such that f(c) is equation to the average value of #f(x)=3x^2#?

How do you find the point c in the interval #0<=x<=3# such that f(c) is equation to the average value of #f(x)=3x^2#?