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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)=2x f(x)=2-x and g(x)=x2 g(x)=x^{2} .

Full solution

Q. Find the area enclosed by the graphs of f(x)=2x f(x)=2-x and g(x)=x2 g(x)=x^{2} .
  1. Identify Intersection Points: Identify the points of intersection between f(x)=2xf(x) = 2 - x and g(x)=x2g(x) = x^2. Set 2x=x22 - x = x^2. Rearrange to x2+x2=0x^2 + x - 2 = 0. Factorize: (x+2)(x1)=0(x + 2)(x - 1) = 0. Solutions: x=2x = -2, x=1x = 1.
  2. Set Up Integral for Area: Set up the integral to find the area between the curves from x=2x = -2 to x=1x = 1. The area AA is given by the integral from 2-2 to 11 of (2xx2)dx(2 - x - x^2) \, dx.
  3. Calculate Integral for Area: Calculate the integral.\newlineA=21(2xx2)dxA = \int_{-2}^{1} (2 - x - x^2) dx.\newline=[2xx22x33]= [2x - \frac{x^2}{2} - \frac{x^3}{3}] from 2-2 to 11.\newline=(2(1)(1)2/2(1)3/3)(2(2)(2)2/2(2)3/3)= (2(1) - (1)^2/2 - (1)^3/3) - (2(-2) - (-2)^2/2 - (-2)^3/3).\newline=(21/21/3)(42(8/3))= (2 - 1/2 - 1/3) - (-4 - 2 - (-8/3)).\newline=(3/21/3)(6+8/3)= (3/2 - 1/3) - (-6 + 8/3).\newline=(9/62/6)(18/3+8/3)= (9/6 - 2/6) - (-18/3 + 8/3).\newline=7/6(10/3)= 7/6 - (-10/3).\newline=7/6+10/3= 7/6 + 10/3.\newline=[2xx22x33]= [2x - \frac{x^2}{2} - \frac{x^3}{3}]00.\newline=[2xx22x33]= [2x - \frac{x^2}{2} - \frac{x^3}{3}]11.\newline=[2xx22x33]= [2x - \frac{x^2}{2} - \frac{x^3}{3}]22.

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