Let f(x)=x-4 and g(x)=(x-4)^(3)". " Find the sum of the areas enclosed by the graphs of f and g between x=3 and x=5. Use a graphing calculator and round your answer to three decimal places.

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Let  f(x)=x-4 and  g(x)=(x-4)^(3)". " Find the sum of the areas enclosed by the graphs of  f and  g between  x=3 and  x=5. Use a graphing calculator and round your answer to three decimal places.

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Understand the Problem: First, we need to understand the problem. We are asked to find the sum of the areas between two curves, f(x) and g(x), from x = 3 to x = 5. To do this, we will calculate the definite integral of the absolute value of the difference between the two functions over the interval [3, 5].Set Up Integral: Let's set up the integral to find the area between the two curves. The area A can be found by integrating the absolute value of the difference between f(x) and g(x) from x = 3 to x = 5:  A = ∫ from 3 to 5 of |f(x) - g(x)| dxDetermine Top Function: We need to determine the function that is on top (greater y-value) over the interval [3, 5]. Since f(x) is a linear function and g(x) is a cubic function with the same root at x = 4, we can expect g(x) to be below f(x) near the root. However, we should verify this by evaluating the functions at a point in the interval, such as x = 3.5.  f(3.5) = 3.5 - 4 = -0.5 g(3.5) = (3.5 - 4)^3 = -0.5^3 = -0.125  Since f(3.5) > g(3.5), f(x) is above g(x) in the interval [3, 5].Write Integral: Now we can write the integral without the absolute value, as we know f(x) is above g(x):  A = ∫ from 3 to 5 of (f(x) - g(x)) dx A = ∫ from 3 to 5 of ((x - 4) - (x - 4)^3) dxCalculate Integral: Let's calculate the integral:  A = ∫ from 3 to 5 of (x - 4 - (x - 4)^3) dx A = [0.5x^2 - 4x - (1/4)x^4 + 4x^3 - 16x^2 + 64x - 64] from 3 to 5  Now we need to evaluate this antiderivative at the bounds x = 5 and x = 3 and subtract the results.Evaluate at x=5: Evaluating the antiderivative at x = 5:  A(5) = 0.5(5)^2 - 4(5) - (1/4)(5)^4 + 4(5)^3 - 16(5)^2 + 64(5) - 64 A(5) = 0.5(25) - 20 - (1/4)(625) + 4(125) - 16(25) + 320 - 64 A(5) = 12.5 - 20 - 156.25 + 500 - 400 + 320 - 64 A(5) = 12.5 - 20 - 156.25 + 500 - 400 + 320 - 64 A(5) = 142.5  Evaluating the antiderivative at x = 3:  A(3) = 0.5(3)^2 - 4(3) - (1/4)(3)^4 + 4(3)^3 - 16(3)^2 + 64(3) - 64 A(3) = 0.5(9) - 12 - (1/4)(81) + 4(27) - 16(9) + 192 - 64 A(3) = 4.5 - 12 - 20.25 + 108 - 144 + 192 - 64 A(3) = 4.5 - 12 - 20.25 + 108 - 144 + 192 - 64 A(3) = 64.25  Now we subtract A(3) from A(5) to find the area:  Area = A(5) - A(3) Area = 142.5 - 64.25 Area = 78.25  However, this calculation is incorrect because the antiderivative was not computed correctly. There is a math error in the antiderivative expression and its evaluation.

Let $f(x)=x-4$ and $g(x)=(x-4)^{3} \text {. }$ $\newline$ Find the sum of the areas enclosed by the graphs of $f$ and $g$ between $x=3$ and $x=5$ . $\newline$ Use a graphing calculator and round your answer to three decimal places.

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Let f(x)=x−4 f(x)=x-4 f(x)=x−4 and g(x)=(x−4)3. g(x)=(x-4)^{3} \text {. } g(x)=(x−4)3. \newlineFind the sum of the areas enclosed by the graphs of f f f and g g g between x=3 x=3 x=3 and x=5 x=5 x=5.\newlineUse a graphing calculator and round your answer to three decimal places.

Let $f(x)=x-4$ and $g(x)=(x-4)^{3} \text {. }$ $\newline$ Find the sum of the areas enclosed by the graphs of $f$ and $g$ between $x=3$ and $x=5$ . $\newline$ Use a graphing calculator and round your answer to three decimal places.