The function f is given by f(x)=3x^2+1. What is the average value of f over the closed interval [1,3]?
The answer I get is 12, but the book says it's 14. I don't understand why.
wait. 12 is the average slope:
(f(3)-f(1))/(2-1)
The average value is the integral
∫[1,3] f(x) dx
----------------- = 14
2
I also get 12.
Well, it seems like there's a discrepancy between your answer and the book's answer. But hey, I'm here to make you smile, not to solve math mysteries!
To find the average value of a function over an interval, we need to evaluate the definite integral of the function over that interval and then divide it by the length of the interval.
So, in this case, we need to evaluate the definite integral of f(x)=3x^2+1 over the interval [1,3] and then divide it by the length of the interval, which is 2.
Let's do the math:
∫[1,3] (3x^2+1) dx = [x^3 + x] from 1 to 3
Plugging in the upper and lower limits:
[3^3 + 3] - [1^3 + 1] = 27 + 3 - 1 - 1 = 28.
Now, divide this by the length of the interval, which is 2:
28 / 2 = 14.
So, the average value of f over the closed interval [1,3] is indeed 14, not 12.
I hope this clears up the confusion for you. Ah, math, always finding a way to keep us entertained!
To find the average value of a function over a closed interval [a, b], you need to evaluate the definite integral of the function over that interval and then divide the result by the width of the interval.
In this case, the function f(x) is given by f(x) = 3x^2 + 1, and the closed interval is [1, 3].
Step 1: Evaluate the definite integral of f(x) over the interval [1, 3]
The definite integral of f(x) over the interval [1, 3] is given by:
∫[1,3] (3x^2 + 1) dx
To evaluate this integral, you need to find the antiderivative of the function f(x) and then apply the Fundamental Theorem of Calculus:
F(x) = x^3 + x
Using the Fundamental Theorem of Calculus, the definite integral of f(x) over [1, 3] is:
∫[1,3] (3x^2 + 1) dx = F(3) - F(1)
= (3^3 + 3) - (1^3 + 1)
= 27 + 3 - 1 - 1
= 29
Step 2: Divide the result by the width of the interval [1, 3]
The width of the interval [1, 3] is given by b - a, which in this case is:
3 - 1 = 2
So, the average value of f over the closed interval [1, 3] is:
Average = (29) / (2) = 14.5
Therefore, the correct answer is 14.5, not 14.
To find the average value of a function over an interval, you need to calculate the definite integral of the function over that interval and divide the result by the width of the interval.
In this case, the function f(x) is given by f(x) = 3x^2 + 1, and the interval is [1,3]. The width of the interval is 3 - 1 = 2.
To find the definite integral of f(x) over the interval [1,3], you can use the power rule for integration. The power rule states that the integral of x^n with respect to x is (1/(n+1)) * x^(n+1) + C, where C is the constant of integration. Applying this rule to the function f(x) = 3x^2 + 1, we get:
∫[1,3] (3x^2 + 1) dx = (1/3) * x^3 + x ∣[1,3]
Now, evaluate the integral at the upper and lower bounds of the interval:
(1/3) * 3^3 + 3 - [(1/3) * 1^3 + 1]
= (1/3) * 27 + 3 - (1/3) * 1 + 1
= 9 + 3 - 1/3 + 1
= 13 + 2/3.
Finally, divide the result by the width of the interval:
(13 + 2/3) / 2 = 14/2 + 2/3 = 7 + 2/3 = 7.6667 (rounded to four decimal places).
Therefore, the average value of f over the closed interval [1,3] is approximately 7.6667. If the book says the answer is 14, it may be a typo or there might be some other misunderstanding. Double-check the book or consult with your instructor to clarify the correct answer.