Application of Integrals: Speed-Solving Techniques (6)

Substitute $y = 2x - 4$ into $y^2 = 4x$:

$(2x - 4)^2 = 4x \implies 4x^2 - 16x + 16 = 4x$

$4x^2 - 20x + 16 = 0 \implies x^2 - 5x + 4 = 0$

$(x - 1)(x - 4) = 0 \implies x = 1, 4$

Corresponding $y$-values: $y = -2$ and $y = 4$.

Integrating with respect to $y$ avoids the parabola’s “two halves” issue. From $y^2 = 4x$, $x = y^2/4$. From $y = 2x - 4$, $x = (y + 4)/2$.

The line $x_{\text{line}} = (y + 4)/2$ lies to the right of the parabola $x_{\text{parab}} = y^2/4$ for $y \in [-2, 4]$.

$A = \int_{-2}^{4} \left[\frac{y + 4}{2} - \frac{y^2}{4}\right] dy$

$A = \left[\frac{y^2}{4} + 2y - \frac{y^3}{12}\right]_{-2}^{4}$

At $y = 4$: $\frac{16}{4} + 8 - \frac{64}{12} = 4 + 8 - \frac{16}{3} = 12 - \frac{16}{3} = \frac{20}{3}$.

At $y = -2$: $\frac{4}{4} - 4 + \frac{8}{12} = 1 - 4 + \frac{2}{3} = -\frac{7}{3}$.

$A = \frac{20}{3} - \left(-\frac{7}{3}\right) = \frac{27}{3} = 9$

Final: Area = 9 square units.