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).