The Average value of f(x)=25−x^2 on the interval [0,1] is 74/3
Find a value c in the interval [0,1] such that f(c) is equal to the average value?
Well, well, well. Looks like we have a math problem on our hands. Finding the value of c where f(c) is equal to the average value of a function? That's quite a task! I'm not sure if I can find the answer, but I can definitely clown around and make you smile while you try to solve it!
Let's see, if f(x) = 25 - x^2 and the average value is 74/3, we want to find c where f(c) = 74/3. So, let's set up the equation:
25 - c^2 = 74/3
Now, we can solve this algebraic equation to find the value of c. Put on your thinking cap and give it a go! And don't worry, even if you don't find the answer, I'll be here to cheer you up!
To find a value c in the interval [0,1] such that f(c) is equal to the average value, we need to solve the equation f(c) = 74/3.
The equation f(c) = 74/3 represents the average value of f(x) on the interval [0,1].
The function f(x) = 25 - x^2.
Substituting c for x in the function f(x), we have f(c) = 25 - c^2.
Now we can set up the equation: 25 - c^2 = 74/3.
To solve this equation, we'll first multiply both sides by 3 to eliminate the fraction: 3(25 - c^2) = 74.
Expanding the left side, we have: 75 - 3c^2 = 74.
Next, subtract 75 from both sides: -3c^2 = -1.
Divide both sides by -3 to isolate c^2: c^2 = 1/3.
Take the square root of both sides: c = ±√(1/3).
Since we are looking for a value c in the interval [0,1], the negative value is not valid. Therefore, c = √(1/3).
Approximating the square root of 1/3 to two decimal places, c ≈ 0.58.
Thus, a value c in the interval [0,1] such that f(c) is equal to the average value 74/3 is approximately 0.58.
To find a value c in the interval [0,1] such that f(c) is equal to the average value, first, we need to find the average value of f(x) on the interval [0,1].
The average value of a function f(x) on the interval [a,b] is given by the formula:
Avg = (1/(b-a)) * ∫[a,b] f(x) dx
In this case, f(x) = 25 - x^2 and the interval is [0,1]. Therefore, the average value can be calculated as follows:
Avg = (1/(1-0)) * ∫[0,1] (25 - x^2) dx
Simplifying the integral:
Avg = (1/1) * [25x - (1/3)x^3] evaluated from 0 to 1
Avg = (1/1) * [(25(1) - (1/3)(1)^3) - (25(0) - (1/3)(0)^3)]
Avg = 25 - (1/3)
Avg = 74/3
So, the average value of f(x) on the interval [0,1] is 74/3.
Now, we need to find a value c in the interval [0,1] such that f(c) is equal to the average value.
To do that, we set f(c) equal to the average value and solve for c:
25 - c^2 = 74/3
Multiplying through by 3 to get rid of the fraction:
75 - 3c^2 = 74
Rearranging the equation:
3c^2 = 75 - 74
3c^2 = 1
Dividing through by 3:
c^2 = 1/3
Taking the square root of both sides:
c = ± √(1/3)
Since c is in the interval [0,1], we take the positive value:
c = √(1/3)
Therefore, a value c approximately equal to √(1/3) in the interval [0,1] will satisfy the equation f(c) = 74/3.
the average value is
(∫[0,1] (25-x^2) dx)/(1-0)
or, since the area of the semicircle is 25/4 π. the average value is, as always,
area / width