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Find the area enclosed by the graphs of
f(x)=2-x and
g(x)=x^(2)

Find the area enclosed by the graphs of $f(x)=2-x$ and $g(x)=x^{2}$

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Domain and range of quadratic functions: equations

Full solution

Q. Find the area enclosed by the graphs of

f(x)=2-x

and

g(x)=x^{2}

Identify Intersection Points: Identify the points of intersection between $f(x) = 2 - x$ and $g(x) = x^2$ . Set $2 - x = x^2$ . $x^2 + x - 2 = 0$ . Factorize: $(x + 2)(x - 1) = 0$ . $x = -2$ , $x = 1$ .
Set Up Integral: Set up the integral to find the area between the curves from $x = -2$ to $x = 1$ . The area $A$ is given by the integral from $-2$ to $1$ of $(2 - x - x^2) \, dx$ .
Calculate Integral: Calculate the integral. $\newline$ $A = \int_{-2}^{1} (2 - x - x^2) dx.$ $\newline$ $= [2x - \frac{x^2}{2} - \frac{x^3}{3}] \text{ from } -2 \text{ to } 1.$ $\newline$ $= (2(1) - (1)^2/2 - (1)^3/3) - (2(-2) - (-2)^2/2 - (-2)^3/3).$ $\newline$ $= (2 - 1/2 - 1/3) - (-4 - 2 - (-8/3)).$ $\newline$ $= (3/2 - 1/3) - (-4 - 2 + 8/3).$ $\newline$ $= (9/6 - 2/6) - (-18/3 - 6/3 + 8/3).$ $\newline$ $= 7/6 - (-16/3).$ $\newline$ $= 7/6 + 16/3.$ $\newline$ $= 7/6 + 32/6.$ $\newline$ $= 39/6.$ $\newline$ $= [2x - \frac{x^2}{2} - \frac{x^3}{3}] \text{ from } -2 \text{ to } 1.$ $0$

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