What is the average value of a function #y=6/x# on the interval #[1,e]#?

Question

Ezekiel Alexander · Accepted Answer

#6/(e-1)approx3.4918602412...#

The average value of the function #f(x)# on the interval #[a,b]# can be expressed as:

#1/(b-a)int_a^bf(x)dx#

So, where the function is #f(x)=6/x# and the interval is #[1,e]#, the average value of the function is:

#1/(e-1)int_1^e 6/xdx#

The #6# can be brought out of the integrand:

#=6/(e-1)int_1^e 1/xdx#

Note that since the derivative of #ln(x)# is #1/x#, the integral of #1/x# is #ln(x)#.

#=6/(e-1)[ln(x)]_1^e#

Now evaluating:

#=6/(e-1)[ln(e)-ln(1)]#

#=6/(e-1)(1-0)#

#=6/(e-1)#

Axel Alvarez · Answer

undefined To find the average value of the function $ y = \frac{6}{x} $ on the interval $[1,e]$, we need to evaluate the definite integral of the function over that interval and then divide by the length of the interval.

The integral of $ y = \frac{6}{x} $ with respect to $ x $ is $ 6\ln(x) $.

So, the average value $ \bar{y} $ of the function on the interval $[1,e]$ is:

$$ \bar{y} = \frac{1}{e-1} \int_{1}^{e} \frac{6}{x} \, dx = \frac{1}{e-1} \left[ 6\ln(x) \right]_{1}^{e} $$

$$ = \frac{1}{e-1} \left[ 6\ln(e) - 6\ln(1) \right] $$

$$ = \frac{1}{e-1} \left[ 6 - 0 \right] $$

$$ = \frac{6}{e-1} $$

So, the average value of the function $ y = \frac{6}{x} $ on the interval $[1,e]$ is $ \frac{6}{e-1} $.