Area Between two Curves

Examples

Find the equation of the intersection between $y = x^{2}$ and $y = 2 x - x^{2}$ .

\begin{aligned} A & = \int (x^{2} - (2 x - x^{2})) d x \\ = 2 \int (x^{2} - x) d x \\ = \frac{2}{3} x^{3} - x^{2} \end{aligned}

Find the area of the intersection between $y = x - 1$ and $y^{2} = 2 x + 6$ :
With respect to $y$ .
With respect to $x$ .

\begin{aligned} y & = x - 1 \\ x & = y + 1 \\ y^{2} & = 2 x + 6 \\ x & = \frac{1}{2} y^{2} - 3 \end{aligned}

Intersecting points:

\begin{aligned} y + 1 & = \frac{1}{2} y^{2} - 3 \\ y^{2} - 2 y - 8 & = 0 \\ (y - 4) (y + 2) & = 0 \\ y & = 4 or - 2 \\ x_{1} & = 4 + 1 = 5 \\ x_{2} & = - 2 + 1 = - 1 \end{aligned}

Area:

\begin{aligned} A & = {[\int (y + 1 - \frac{1}{2} y^{2} + 3) d y]}_{- 2}^{4} \\ = {[- \frac{1}{6} y^{3} + \frac{1}{2} y^{2} + 4 y + c]}_{- 2}^{4} \\ = - \frac{1}{6} 4^{3} + \frac{1}{2} 4^{2} + 4 \cdot 4 + \frac{1}{6} (- 2)^{3} - \frac{1}{2} (- 2)^{2} - 4 (- 2) \\ = 18 \end{aligned}

\begin{aligned} A & = {[\int (\sqrt{2 x + 6} + \sqrt{2 x + 6}) d x]}_{- 3}^{- 1} + [\sqrt{2 x + 6} - x + 1]_{- 1}^{5} \\ = {[\frac{2}{3} (2 x + 6)^{\frac{3}{2}} + c]}_{- 3}^{- 1} + {[\frac{1}{3} (2 x + 6)^{\frac{3}{2}} - \frac{1}{2} x^{2} + x + c]}_{- 1}^{5} \\ = (\frac{2}{3} (2 \cdot - 1 + 6)^{\frac{3}{2}} - \frac{2}{3} (2 \cdot - 3 + 6)^{\frac{3}{2}}) + (\frac{1}{3} (2 \cdot 5 + 6)^{\frac{3}{2}} - \frac{1}{2} 5^{2} + 5 - \frac{1}{3} (2 \cdot - 1 + 6)^{\frac{3}{2}} + \frac{1}{2} (- 1)^{2} + 1) \\ = 4.666 \dots \end{aligned}

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