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Find the average value of the function 
f(x)=(12)/(9-x) from 
x=1 to 
x=7. Express your answer as a constant times 
ln 2.
Answer: 
◻ln 2

Find the average value of the function f(x)=129x f(x)=\frac{12}{9-x} from x=1 x=1 to x=7 x=7 . Express your answer as a constant times ln2 \ln 2 .\newlineAnswer: ln2 \square\ln 2

Full solution

Q. Find the average value of the function f(x)=129x f(x)=\frac{12}{9-x} from x=1 x=1 to x=7 x=7 . Express your answer as a constant times ln2 \ln 2 .\newlineAnswer: ln2 \square\ln 2
  1. Set up integral: To find the average value of a continuous function f(x)f(x) on the interval [a,b][a, b], we use the formula for the average value of a function:\newlineAverage value = (1/(ba))×abf(x)dx(1/(b-a)) \times \int_{a}^{b} f(x) \, dx\newlineHere, a=1a = 1 and b=7b = 7, so we need to calculate the integral of f(x)=(12)/(9x)f(x) = (12)/(9-x) from x=1x=1 to x=7x=7 and then multiply by 1/(71)1/(7-1).
  2. Perform substitution: First, let's set up the integral:\newline17129xdx\int_{1}^{7} \frac{12}{9-x} \, dx\newlineTo integrate this function, we can use a substitution method. Let u=9xu = 9 - x, which means du=dxdu = -dx. We also need to change the limits of integration according to the substitution.
  3. Evaluate new limits: When x=1x = 1, u=91=8u = 9 - 1 = 8. When x=7x = 7, u=97=2u = 9 - 7 = 2. So the new limits of integration are from u=8u = 8 to u=2u = 2. The integral becomes:\newline8212u(du)\int_{8}^{2} \frac{12}{u} \cdot (-du)\newlineNotice that we have a negative sign because du=dxdu = -dx.
  4. Integrate function: We can pull out the constants and reverse the limits of integration to get rid of the negative sign:\newline12×28(1u)du-12 \times \int_{2}^{8} (\frac{1}{u}) \, du\newlineNow we can integrate 1u\frac{1}{u} with respect to uu.
  5. Evaluate expression: The integral of 1u\frac{1}{u} du is lnu\ln|u|, so we have:\newline12×[lnu]-12 \times [\ln|u|] from u=2u = 2 to u=8u = 8\newlineNow we need to evaluate this expression using the limits of integration.
  6. Apply average value formula: Plugging in the limits, we get:\newline12×(ln8ln2)-12 \times (\ln|8| - \ln|2|)\newlineSince ln8=ln(23)=3ln(2)\ln|8| = \ln(2^3) = 3\ln(2) and ln2=ln(2)\ln|2| = \ln(2), the expression simplifies to:\newline12×(3ln(2)ln(2))=12×(2ln(2))-12 \times (3\ln(2) - \ln(2)) = -12 \times (2\ln(2))
  7. Correct mistake: Now we have the integral evaluated, but we need to remember to multiply by 1/(ba)1/(b-a) to find the average value. In this case, ba=71=6b-a = 7-1 = 6. So we multiply 12×(2ln(2))-12 \times (2\ln(2)) by 1/61/6: \newlineAverage value = (1/6)×12×(2ln(2))=4ln(2)(1/6) \times -12 \times (2\ln(2)) = -4\ln(2)
  8. Correct mistake: Now we have the integral evaluated, but we need to remember to multiply by 1(ba)\frac{1}{(b-a)} to find the average value. In this case, ba=71=6b-a = 7-1 = 6. So we multiply 12×(2ln(2))-12 \times (2\ln(2)) by 16\frac{1}{6}:
    Average value = (16)×12×(2ln(2))=4ln(2)(\frac{1}{6}) \times -12 \times (2\ln(2)) = -4\ln(2)However, we made a mistake in the previous step. The average value should be positive because we are looking for the average of a function over an interval, and the function is positive in the given interval. The mistake was in the limits of integration after the substitution. We should have kept the original order and accounted for the negative sign within the integral.
    Let's correct this:
    Average value = (16)×12×(2ln(2))=4ln(2)(\frac{1}{6}) \times 12 \times (2\ln(2)) = 4\ln(2)

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