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Find the average value of the function f(x)=(10)/(8x-3) from x=1 to x=6.

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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).

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