Application of Integrals: Step-by-Step Worked Examples (2)

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

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

$x^2 - 5x + 4 = 0 \implies x = 1 \text{ or } x = 4$

Corresponding $y$ values: $y = -2$ at $x = 1$, $y = 4$ at $x = 4$.

So intersection points are $(1, -2)$ and $(4, 4)$.

Integrating with respect to $y$ is cleaner here because both curves are functions of $y$ on this region:

• Parabola: $x = y^2/4$
• Line: $x = (y+4)/2$

For $-2 \leq y \leq 4$, the line is to the right of the parabola (check by plugging $y = 0$: line gives $x = 2$, parabola gives $x = 0$).

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

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

At $y = 4$: $4 + 8 - 64/12 = 12 - 16/3 = 20/3$.

At $y = -2$: $1 - 4 + 8/12 = -3 + 2/3 = -7/3$.

$A = 20/3 - (-7/3) = 27/3 = 9$