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Find the average value of the function f(x)=(4)/(6-x) from x=7 to x=9. Write your answer as the logarithm of a single number in simplest form. Answer: ln(◻)

Find the average value of the function f(x)=46−x f(x)=\frac{4}{6-x} from x=7 x=7 to x=9 x=9 . Write your answer as the logarithm of a single number in simplest form.\newlineAnswer: ln⁡(□) \ln (\square)

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Q. Find the average value of the function f(x)=46−x f(x)=\frac{4}{6-x} from x=7 x=7 to x=9 x=9 . Write your answer as the logarithm of a single number in simplest form.\newlineAnswer: ln⁡(□) \ln (\square)
  1. Set Up Integral: The average value of a continuous function f(x)f(x) on the interval [a,b][a, b] is given by the formula:\newlineAverage value = (1/(b−a))⋅∫abf(x) dx(1/(b-a)) \cdot \int_{a}^{b} f(x) \, dx\newlineHere, a=7a = 7, b=9b = 9, and f(x)=(4)/(6−x)f(x) = (4)/(6-x).\newlineFirst, we need to set up the integral to find the average value.
  2. Calculate Integral: Now we calculate the integral of f(x)f(x) from x=7x=7 to x=9x=9. \newline∫7946−x dx\int_{7}^{9} \frac{4}{6-x} \, dx\newlineTo integrate this function, we can use a substitution method. Let u=6−xu = 6-x, then du=−dxdu = -dx.
  3. Substitution Method: Substituting uu into the integral, we get:\newline∫7946−xdx=−∫−1−34udu\int_{7}^{9} \frac{4}{6-x} dx = -\int_{-1}^{-3} \frac{4}{u} du\newlineNotice that the limits of integration have also changed according to the substitution u=6−xu = 6-x. When x=7x = 7, u=−1u = -1, and when x=9x = 9, u=−3u = -3.
  4. Evaluate Integral: Now we integrate 4u\frac{4}{u} with respect to uu:∫4udu=4⋅ln⁡∣u∣\int \frac{4}{u} du = 4 \cdot \ln|u| Evaluating this from −1-1 to −3-3 gives us:4⋅ln⁡∣−3∣−4⋅ln⁡∣−1∣4 \cdot \ln|-3| - 4 \cdot \ln|-1|
  5. Simplify Expression: Since ln⁡∣−1∣=ln⁡(1)=0\ln|-1| = \ln(1) = 0, the expression simplifies to:\newline4×ln⁡∣−3∣−4×0=4×ln⁡(3)4 \times \ln|-3| - 4 \times 0 = 4 \times \ln(3)
  6. Find Average Value: Now we need to divide this result by (b−a)(b-a) to find the average value:\newlineAverage value = 1(9−7)×4×ln⁡(3)=12×4×ln⁡(3)=2×ln⁡(3)\frac{1}{(9-7)} \times 4 \times \ln(3) = \frac{1}{2} \times 4 \times \ln(3) = 2 \times \ln(3)
  7. Final Result: The average value of the function f(x)f(x) from x=7x=7 to x=9x=9 is therefore 2⋅ln⁡(3)2 \cdot \ln(3). This is the logarithm of a single number, as required.

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