Answer to Question #201029 in Calculus for Xcon

Question

Find the area between two curves: y^2 = 2x + 3 ; y = x

Answer 1

$$y^2=2x+3=>x=\dfrac{1}{2}y^2-\dfrac{3}{2}$$

$$y=x=>y=\dfrac{1}{2}y^2-\dfrac{3}{2}$$

$$y^2-2y-3=0$$

$$(y+1)(y-3)=0$$

$Point (-1,-1), Point(3,3)$

$$Area=\displaystyle\int_{-1}^{3}|y-(\dfrac{1}{2}y^2-\dfrac{3}{2})|dy$$

$$=\displaystyle\int_{-1}^{3}(y-\dfrac{1}{2}y^2+\dfrac{3}{2})dy$$

$$=[\dfrac{1}{2}y^2-\dfrac{1}{6}y^3+\dfrac{3}{2}y]\begin{matrix} 3 \\ -1 \end{matrix}$$

$$=\dfrac{9}{2}-\dfrac{27}{6}+\dfrac{9}{2}-\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{3}{2}$$

$$=\dfrac{16}{3} (units^2)$$

Area is $\dfrac{16}{3}$ square units.