To find the average value of the function f(x) = 10/(8x - 3) from x=1 to x=6, we must use the formula for the average value of a function on the interval [a, b]:
The formula is: (1/(b-a)) ∫ from a to b [f(x) dx]
Here, our function f(x) is 10/(8x - 3), a (lower limit of the interval) is 1, and b (upper limit of the interval) is 6.
Substitute f(x), a, and b into the formula:
average value = (1/(6-1)) ∫ from 1 to 6 [10/(8x - 3) dx]
Simplify this to:
average value = (1/5) ∫ from 1 to 6 [10/(8x - 3) dx]
This is a basic integral, but might seem complex due to the presence of x in the denominator.
To solve the integral, we apply the rule of integration for functions that resemble 1/x, which states that the antiderivative of 1/x is log |x|.
So, ∫[10/(8x - 3) dx] = 1.25 * [log |8x-3|], note that 10/8 = 1.25
Now, we evaluate this at the limits of 6 and 1.
= 1.25 * [log |8*6-3| - log |8*1-3|]
Keep in mind the logarithm property that allows us to express a difference of logs as a quotient within one log:
= 1.25 * log|(8*6-3)/(8*1-3)|
Plug in the x-values:
= 1.25 * log|(48-3)/(8-3)|
Simplify inside the absolute bars:
= 1.25 * log|45/5|
Finally, simplify inside the log:
= 1.25 * log(9)
But, remember, the original expression for the average value included a factor of 1/5, so we carry that factor back in:
average value = (1/5) * 1.25 * log(9)
Finally, simplify to get:
average value = -0.25*log(5) + 0.25*log(45)
And there you have it! The average value of the function f(x) = 10/(8x - 3) from x=1 to x=6 is -0.25*log(5) + 0.25*log(45).