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Let
f(x)=(1)/(2)x^(3)+(1)/(2)x^(2)-x and
g(x)=x^(2)+2x.
Find the sum of the areas enclosed by the graphs of
f and
g between
x=-2 and
x=3.
Use a graphing calculator and round your answer to three decimal places.

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

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Complex conjugate theorem

Full solution

Q. Let

f(x)=\frac{1}{2} x^{3}+\frac{1}{2} x^{2}-x

and

g(x)=x^{2}+2 x

.

\newline

Find the sum of the areas enclosed by the graphs of

f

and

g

between

x=-2

and

x=3

.

\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 $x=-2$ to $x=3$ . The area between the curves can be found by integrating the absolute value of the difference between the functions over the given interval.
Set up the integral: Set up the integral to find the area. $\newline$ The area $A$ between the two curves from $x=a$ to $x=b$ is given by the integral of the absolute value of the difference of the functions: $\newline$ $A = \int_{a}^{b} |f(x) - g(x)| \, dx$ $\newline$ In this case, $a = -2$ and $b = 3$ .
Calculate the difference: Calculate the difference between the functions. $\newline$ Subtract $g(x)$ from $f(x)$ to find the integrand: $\newline$ $h(x) = f(x) - g(x) = (\frac{1}{2})x^3 + (\frac{1}{2})x^2 - x - (x^2 + 2x)$ $\newline$ Simplify $h(x)$ : $\newline$ $h(x) = (\frac{1}{2})x^3 - (\frac{1}{2})x^2 - 3x$
Find the points of intersection: Find the points of intersection. $\newline$ To find the points where the graphs of $f(x)$ and $g(x)$ intersect, we set $f(x)$ equal to $g(x)$ and solve for $x$ : $\newline$ $(1/2)x^3 + (1/2)x^2 - x = x^2 + 2x$ $\newline$ $(1/2)x^3 - (3/2)x^2 - 3x = 0$ $\newline$ $x(x^2 - 3x - 6) = 0$ $\newline$ Solve for $x$ to find the points of intersection.
Solve the equation for $x$ : Solve the equation for $x$ . $x(x - 3)(x + 2) = 0$ The solutions are $x = 0$ , $x = 3$ , and $x = -2$ .These are the points of intersection, and they match our interval from $x = -2$ to $x = 3$ .
Integrate the absolute value: Integrate the absolute value of $h(x)$ from $x = -2$ to $x = 3$ . Since we have the points of intersection, we can integrate $h(x)$ without taking the absolute value, as we know the function that is on top for each interval. We need to split the integral at the points of intersection if the top function changes. $A = \int_{-2}^{0} h(x) \, dx + \int_{0}^{3} h(x) \, dx$
Use a graphing calculator: Use a graphing calculator to evaluate the integrals. $\newline$ Since the problem asks us to use a graphing calculator, we will input the function $h(x)$ and use the calculator's integration function to find the area from $x = -2$ to $x = 0$ and from $x = 0$ to $x = 3$ .
Add the areas: Add the areas from each interval. $\newline$ Let $A_1$ be the area from $x = -2$ to $x = 0$ , and $A_2$ be the area from $x = 0$ to $x = 3$ . The total area $A$ is $A_1 + A_2$ . $\newline$ Use the graphing calculator to find $A_1$ and $A_2$ , and then add them together.
Round the final answer: Round the final answer to three decimal places. $\newline$ After calculating $A_1$ and $A_2$ , add them to get the total area $A$ . Round $A$ to three decimal places as instructed.

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