What is the area of the region between the graphs of f(x)=(4)/(x) , g(x)=4, and x=4 ? Choose 1 answer: (A) 24-4ln(4) (B) 12-4ln(3) (C) 12-4ln(4) (D) 24-4ln(3)

Question

What is the area of the region between the graphs of  f(x)=(4)/(x) , g(x)=4, and  x=4 ? Choose 1 answer: (A)  24-4ln(4) (B)  12-4ln(3) (C)  12-4ln(4) (D)  24-4ln(3)

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Set Up Integral: To find the area between the curves, we need to set up an integral from the left boundary to the right boundary of the region. The left boundary is where the two functions intersect, which we need to find.Find Intersection Point: To find the intersection point of f(x) and g(x), we set f(x) equal to g(x) and solve for x: (4)/(x) = 4 Multiplying both sides by x to clear the fraction, we get: 4 = 4x Dividing both sides by 4, we get: 1 = x So, the left boundary is x = 1.Calculate Area Integral: The area A between the curves from x = 1 to x = 4 is given by the integral of the top function minus the bottom function. Since g(x) = 4 is above f(x) = (4)/(x) in this interval, we have: A = ∫ from 1 to 4 of (g(x) - f(x)) dx A = ∫ from 1 to 4 of (4 - (4)/(x)) dxEvaluate Integral: We can now evaluate the integral: A = ∫ from 1 to 4 of (4 - (4)/(x)) dx A = ∫ from 1 to 4 of 4 dx - ∫ from 1 to 4 of (4)/(x) dx A = [4x] from 1 to 4 - [4ln|x|] from 1 to 4Plug in Limits: Plugging in the limits of integration, we get: A = (4*4 - 4*1) - (4ln|4| - 4ln|1|) A = (16 - 4) - (4ln(4) - 4ln(1)) Since ln(1) = 0, we simplify further: A = 12 - 4ln(4)Compare with Answer: Comparing our result with the answer choices, we see that our answer matches option (C).

What is the area of the region between the graphs of $f(x)=\frac{4}{x}$ , $g(x)=4$ , and $x=4$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $24-4 \ln (4)$ $\newline$ (B) $12-4 \ln (3)$ $\newline$ (C) $12-4 \ln (4)$ $\newline$ (D) $24-4 \ln (3)$

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