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Find the average value of the function 
f(x)=(4)/(2x-11) from 
x=6 to 
x=10. Write your answer as the logarithm of a single number in simplest form.
Answer: 
ln(◻)

Find the average value of the function f(x)=42x11 f(x)=\frac{4}{2 x-11} from x=6 x=6 to x=10 x=10 . Write your answer as the logarithm of a single number in simplest form.\newlineAnswer: ln() \ln (\square)

Full solution

Q. Find the average value of the function f(x)=42x11 f(x)=\frac{4}{2 x-11} from x=6 x=6 to x=10 x=10 . Write your answer as the logarithm of a single number in simplest form.\newlineAnswer: ln() \ln (\square)
  1. Calculate Interval Length: To find the average value of the function f(x)=42x11f(x) = \frac{4}{2x - 11} over the interval [6,10][6, 10], we need to integrate the function over this interval and then divide by the length of the interval.\newlineThe average value formula is given by:\newlineAverage value = 1(ba)×abf(x)dx\frac{1}{(b - a)} \times \int_{a}^{b} f(x) \, dx\newlineHere, a=6a = 6 and b=10b = 10.
  2. Set Up Integral: First, we calculate the length of the interval [6,10][6, 10], which is bab - a.\newlineLength of interval = 106=410 - 6 = 4
  3. Perform Substitution: Next, we set up the integral of f(x)f(x) from 66 to 1010.61042x11dx\int_{6}^{10} \frac{4}{2x - 11} \, dxWe will need to use a substitution to solve this integral.Let u=2x11u = 2x - 11, then du=2dxdu = 2 \, dx.
  4. Change Limits: We need to change the limits of integration to match our substitution.\newlineWhen x=6x = 6, u=2(6)11=1u = 2(6) - 11 = 1.\newlineWhen x=10x = 10, u=2(10)11=9u = 2(10) - 11 = 9.\newlineNow we can rewrite the integral in terms of uu.\newline194u12du\int_{1}^{9} \frac{4}{u} \cdot \frac{1}{2} \, du\newlineThe 12\frac{1}{2} comes from the fact that du=2dxdu = 2 dx, so dx=12dudx = \frac{1}{2} du.
  5. Simplify Integral: Simplify the integral.\newline\int_{1}^{9} \frac{4}{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \cdot \int_{1}^{9} \frac{4}{u} \, du\(\newline= \frac{1}{2} \cdot \int_{1}^{9} 4u^{-1} \, du\newline= \frac{1}{2} \cdot 4 \cdot \int_{1}^{9} u^{-1} \, du\newline= 2 \cdot \int_{1}^{9} u^{-1} \, du\)
  6. Integrate u1u^{-1}: Now we integrate u1u^{-1} with respect to uu from 11 to 99.u1du\int u^{-1} \, du is the natural logarithm function, ln(u)\ln(u). So, 2×19u1du=2×[ln(u)]192 \times \int_{1}^{9} u^{-1} \, du = 2 \times [\ln(u)]_{1}^{9} = 2×(ln(9)ln(1))2 \times (\ln(9) - \ln(1))
  7. Simplify Expression: Since ln(1)=0\ln(1) = 0, we simplify the expression.2×(ln(9)ln(1))=2×ln(9)2 \times (\ln(9) - \ln(1)) = 2 \times \ln(9)
  8. Find Average Value: Now we divide by the length of the interval to find the average value.\newlineAverage value = (14)×2×ln(9)(\frac{1}{4}) \times 2 \times \ln(9)\newline= (12)×ln(9)(\frac{1}{2}) \times \ln(9)
  9. Simplify ln(9)\ln(9): We can simplify ln(9)\ln(9) since 99 is a perfect square.\newlineln(9)=ln(32)=2×ln(3)\ln(9) = \ln(3^2) = 2 \times \ln(3)\newlineSo, the average value becomes:\newlineAverage value = (1/2)×2×ln(3)(1/2) \times 2 \times \ln(3)\newline= ln(3)\ln(3)

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