Answer to Question #193253 in Calculus for April

Question #193253

Find the area between the curves y^2 - y + x = 0 and y^2 = x + 3.


1
Expert's answer
2021-05-19T12:50:39-0400

Find the area between the curves y2y+x=0y^2-y+x=0 and y2=x+3y^2=x+3.

Solution:\textbf{Solution:}

Let x2(y)=yy2x_{2}(y)=y-y^2, x1(y)=y23x_{1}(y)=y^2-3.

Let's draw the graphs of these functions.



We have two points of intersection.

yy2=y232y2y3=0y-y^2=y^2-3 \Rightarrow 2y^2-y-3=0.

y1=1.5x1=0.75y_{1}=1.5 \Rightarrow x_{1}=-0.75.

y2=1x2=2y_{2}=-1 \Rightarrow x_{2}=-2.

Find the area SS.

S=y2y1(x2(y)x1(y))dy.S=\int_{y_2}^{y_1} (x_{2}(y)-x_{1}(y)) dy.


S=11.5((yy2)(y23))dy=11.5(2y2+y+3)dy.S=\int_{-1}^{1.5} ((y-y^2)-(y^2-3)) dy=\int_{-1}^{1.5} (-2y^2+y+3) dy.


S=(2)11.5y2dy+11.5ydy+311.5dy=S=(-2)\int_{-1}^{1.5} y^2 dy+\int_{-1}^{1.5} y dy+3\int_{-1}^{1.5} dy= (2)y3311.5+y2211.5+3y11.5=12524(-2)\left.\frac{y^3}{3}\right|_{-1}^{1.5}+\left.\frac{y^2}{2}\right|_{-1}^{1.5}+3\left.y\right|_{-1}^{1.5}=\frac{125}{24}.

Answer:\textbf{Answer:}

S=12524\boxed{S=\frac{125}{24}}


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