Answer to Question #280867 in Calculus for Venkatesh

Question

Find the area between y=x^2 and x+y-2=0

Answer 1

$$y=x^2$$

Substitute

$$x+x^2-2=0$$

$$(x+2)(x-1)=0$$

$$x_1=-2, x_2=1$$

$Point(-2, 4), \ Point(1,1).$

$$x+y-2=0$$

$$y=-x+2$$

$$A=\displaystyle\int_{-2}^{1}(-x+2-x^2)dx$$

$$=\big[-\dfrac{x^3}{3}+2x-\dfrac{x^2}{2}\big]\begin{matrix} 1 \\ -2 \end{matrix}$$

$$=-\dfrac{1^3}{3}+2(1)-\dfrac{1^2}{2}-(\dfrac{(-2)^3}{3}+2(-2)-\dfrac{(-2)^2}{2})$$

$$=-\dfrac{1}{3}+2-\dfrac{1}{2}-\dfrac{8}{3}+4+2=\dfrac{9}{2}({units}^2)$$

$Area=\dfrac{9}{2}$ square units.