Find the average interval of function 3x^3-12x^2 on the interval [-5,4]?
My answer is 114.75 but when i put it on submission , it tell me that its wrong not sure why kindly help and tell me right answer
I guess maybe you mean the average value of the function in that domain.
integral = (3/4)x^4-4x^3
at -5 = (3/4)625-4(-125) = 968.75
at 4 = (3)64-4(64) = -64
so I get
968.75 + 64 = 1032.75
divide by length of domain, 9
I agree with you, 114.75
I think you subtracted wrong. Try
(f(4)-f(-5))/9 = -114.75
as a quick look at the graph would indicate:
http://www.wolframalpha.com/input/?i=3x%5E3-12x%5E2
I have put both of these answer's but still wrong don't know what to do
well, I get -114.755
so maybe the key wants -114.76 or -114.8 or -115
To find the average interval of a function on a given interval, you need to calculate the definite integral of the function over that interval, and then divide the result by the length of the interval.
Let's work through the steps to find the average interval of the function f(x) = 3x^3 - 12x^2 on the interval [-5, 4].
Step 1: Calculate the definite integral of the function over the interval [-5, 4].
To integrate the function, we need to find its antiderivative. The antiderivative of 3x^3 is (3/4)x^4, and the antiderivative of -12x^2 is -4x^3. Integrating each term, we get:
∫[3x^3 - 12x^2] dx = (3/4)x^4 - 4x^3
Now, evaluate the antiderivative at the upper and lower limits of the interval:
F(4) = (3/4)(4^4) - 4(4^3) = 192 - 256 = -64
F(-5) = (3/4)(-5^4) - 4(-5^3) = 1875 - 500 = 1375
Step 2: Calculate the length of the interval [-5, 4].
The length of the interval is obtained by subtracting the lower limit from the upper limit:
Length = 4 - (-5) = 4 + 5 = 9
Step 3: Calculate the average interval.
The average interval is obtained by dividing the definite integral by the length of the interval:
Average interval = (1/Length) ∫[3x^3 - 12x^2] dx = (1/9)(-64 - 1375) = -1439/9 = -159.88 (approximately)
Therefore, the correct average interval of the function 3x^3 - 12x^2 on the interval [-5, 4] is approximately -159.88.