Find the average value of the function on the given interval. f(x)=6x lnx; [1,e^2]
average value is area under the curve divided by the interval length
what do you get?
Use integration by parts for x lnx
(1/e^2-1)(6x^2 lnx/2-x^2/4)
am I on the right track?
Integrate the function over that interval and divide the integral by e^2 - 1.
Try integration by parts for the integral.
Let lnx = u and dv = 6 x dx
du = dx/x and v = 3 x^2
Integral of f(x) dx =Integral u dv
= uv - Integral v du
= 3 x^2 lnx - Integral of 3x dx
= 3x^2 lnx - (3/2)x^2
so my answer would be 3/2 (3e^4+1)/ (e^2-1)?
That is correct.
To find the average value of a function on a given interval, you need to follow these steps:
1. Integrate the function over the given interval.
2. Divide the result by the length of the interval.
In this case, you have the function f(x) = 6x ln(x) and the interval [1, e^2].
Step 1: Integrate the function over the interval [1, e^2].
To integrate f(x) = 6x ln(x), we need to rewrite it as the product of two functions and apply integration by parts.
Let's consider u = ln(x) and dv = 6x dx.
By applying the product rule, we get du = (1/x) dx and v = 3x^2.
Therefore, using integration by parts, we have:
∫ 6x ln(x) dx = 3x^2 ln(x) - ∫ 3x^2 (1/x) dx
= 3x^2 ln(x) - ∫ 3x dx
= 3x^2 ln(x) - (3/2) x^2 + C
Next, we need to evaluate this expression at the upper and lower limits of the interval [1, e^2].
At x = e^2:
3(e^2)^2 ln(e^2) - (3/2) (e^2)^2 + C
At x = 1:
3(1)^2 ln(1) - (3/2) (1)^2 + C
Step 2: Calculate the difference of those two values and divide by the length of the interval.
The length of the interval [1, e^2] is given by (e^2) - 1.
So, the average value of f(x) on the interval [1, e^2] is:
[(3(e^2)^2 ln(e^2) - (3/2) (e^2)^2 + C) - (3(1)^2 ln(1) - (3/2) (1)^2 + C)] / (e^2 - 1)
Simplifying further based on the known values of ln(e) = 1 and ln(1) = 0, we have:
[(3(e^2)^2 - (3/2) (e^2)^2) - (0 - (3/2))] / (e^2 - 1)
Finally, simplify the expression:
[(3e^4 - (3/2) e^4) + (3/2)] / (e^2 - 1)
=(3/2) (e^4 + 1) / (e^2 - 1)
Therefore, the average value of f(x) = 6x ln(x) on the interval [1, e^2] is (3/2) (e^4 + 1) / (e^2 - 1).