Answer to Question #193253 in Calculus for April

Question

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

Answer 1

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

$\textbf{Solution:}$

Let $x_{2}(y)=y-y^2$, $x_{1}(y)=y^2-3$.

Let's draw the graphs of these functions.

We have two points of intersection.

$y-y^2=y^2-3 \Rightarrow 2y^2-y-3=0$.

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

$y_{2}=-1 \Rightarrow x_{2}=-2$.

Find the area $S$.

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

$S=\int_{-1}^{1.5} ((y-y^2)-(y^2-3)) dy=\int_{-1}^{1.5} (-2y^2+y+3) dy.$

$S=(-2)\int_{-1}^{1.5} y^2 dy+\int_{-1}^{1.5} y dy+3\int_{-1}^{1.5} dy=$ $(-2)\left.\frac{y^3}{3}\right|_{-1}^{1.5}+\left.\frac{y^2}{2}\right|_{-1}^{1.5}+3\left.y\right|_{-1}^{1.5}=\frac{125}{24}$.

$\textbf{Answer:}$

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