Question

The area bounded by the curves $y = (x + 2)^2$, $y=(x-2)^2$ and the line y = 0 is

• A. $\frac{128}{3}$
• B. $\frac{64}{3}$
• C. $\frac{32}{3}$
• D. $\frac{16}{3}$

Answer 1

Answer: (D)

$\frac{16}{3}$

Answer 2

To find the area bounded by the curves $y = (x + 2)^2$, $y=(x-2)^2$, and the line $y = 0$, we first analyze the curves and their intersection points.

1. Analyze the curves:

   • $y = (x + 2)^2$: This is a parabola opening upwards with its vertex at $(-2, 0)$.
   • $y = (x - 2)^2$: This is a parabola opening upwards with its vertex at $(2, 0)$.
   • $y = 0$: This is the x-axis.

2. Find intersection points:

   • Intersection of $y = (x + 2)^2$ and $y = 0$: $(x + 2)^2 = 0 \implies x = -2$. Point: $(-2, 0)$.
   • Intersection of $y = (x - 2)^2$ and $y = 0$: $(x - 2)^2 = 0 \implies x = 2$. Point: $(2, 0)$.
   • Intersection of $y = (x + 2)^2$ and $y = (x - 2)^2$: $(x + 2)^2 = (x - 2)^2$ $x^2 + 4x + 4 = x^2 - 4x + 4$ $8x = 0 \implies x = 0$. Substitute $x = 0$ into either equation: $y = (0+2)^2 = 4$. Point: $(0, 4)$.

3. Sketch the region:

   The two parabolas open upwards. Their vertices are at $(-2, 0)$ and $(2, 0)$ respectively. They intersect at $(0, 4)$. The line $y=0$ is the x-axis. The region bounded by these three curves is enclosed by the parabolas from above and the x-axis from below.

   The region can be visualized as an area symmetric about the y-axis. The total area can be divided into two parts:

   • Area from $x = -2$ to $x = 0$, bounded by $y = (x + 2)^2$ and $y = 0$.
   • Area from $x = 0$ to $x = 2$, bounded by $y = (x - 2)^2$ and $y = 0$.

4. Set up the integral(s):

   The total area $A$ is the sum of these two integrals: $A = \int_{-2}^{0} (x+2)^2 dx + \int_{0}^{2} (x-2)^2 dx$

5. Evaluate the integrals:

   First integral: $\int_{-2}^{0} (x+2)^2 dx$ Let $u = x+2$, so $du = dx$. When $x=-2$, $u=0$. When $x=0$, $u=2$. $\int_{0}^{2} u^2 du = \left[ \frac{u^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}$.

   Second integral: $\int_{0}^{2} (x-2)^2 dx$ Let $v = x-2$, so $dv = dx$. When $x=0$, $v=-2$. When $x=2$, $v=0$. $\int_{-2}^{0} v^2 dv = \left[ \frac{v^3}{3} \right]_{-2}^{0} = \frac{0^3}{3} - \frac{(-2)^3}{3} = 0 - \frac{-8}{3} = \frac{8}{3}$.

6. Calculate total area: $A = \frac{8}{3} + \frac{8}{3} = \frac{16}{3}$.

The area bounded by the curves is $\frac{16}{3}$.