Let $f(x)=8-2 x$ and $g(x)=x^{3}-7 x^{2}+12 x \text {. }$ $\newline$ Find the sum of the areas enclosed by the graphs of $f$ and $g$ between $x=1$ and $x=4$ . $\newline$ Use a graphing calculator and round your answer to three decimal places.

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Algebra 2

Complex conjugate theorem

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Q. Let

f(x)=8-2 x

and

g(x)=x^{3}-7 x^{2}+12 x \text {. }

\newline

Find the sum of the areas enclosed by the graphs of

f

and

g

between

x=1

and

x=4

\newline

Use a graphing calculator and round your answer to three decimal places.

Understand the problem: Understand the problem. $\newline$ We need to find the area between the curves of $f(x)$ and $g(x)$ from $x=1$ to $x=4$ . This is equivalent to integrating the absolute value of the difference between $f(x)$ and $g(x)$ over the interval $[1, 4]$ .
Set up the integral: Set up the integral to find the area between the curves. $\newline$ The area $A$ between the curves is given by the integral from $x=1$ to $x=4$ of the absolute value of the difference between $f(x)$ and $g(x)$ , which is $|f(x) - g(x)|$ . So, we have: $\newline$ $A = \int_{1}^{4} |f(x) - g(x)| \, dx$
Calculate $f(x) - g(x)$ : Calculate $f(x) - g(x)$ . Before we can integrate, we need to find the expression for $f(x) - g(x)$ : $f(x) - g(x) = (8 - 2x) - (x^3 - 7x^2 + 12x) = -x^3 + 7x^2 - 14x + 8$
Determine points of intersection: Determine the points of intersection between $f(x)$ and $g(x)$ . To find the points of intersection, we set $f(x) = g(x)$ : $8 - 2x = x^3 - 7x^2 + 12x$ $0 = x^3 - 7x^2 + 14x - 8$ We can use a graphing calculator to find the points of intersection within the interval $[1, 4]$ .
Use graphing calculator: Use a graphing calculator to find the points of intersection. $\newline$ After plotting the functions on a graphing calculator, we find that the functions intersect at $x=1$ and $x=4$ . Since these are the limits of our integral, we do not need to split the integral into multiple parts.
Integrate to find area: Integrate the function to find the area. $\newline$ We can now integrate the function from $x=1$ to $x=4$ . Since we are looking for the area, we take the absolute value of the function: $\newline$ $A = \int_{1}^{4} |-x^3 + 7x^2 - 14x + 8| \, dx$ $\newline$ We can use a graphing calculator to evaluate this integral.
Evaluate integral: Evaluate the integral using a graphing calculator. $\newline$ Using the graphing calculator, we find the integral from $x=1$ to $x=4$ of $|-x^3 + 7x^2 - 14x + 8|$ dx to be approximately $18.667$ .