Understanding the Area Bounded by |y| = 1 – x²

The question asks for the area enclosed by the curve defined by the equation |y| = 1 – x². To find the area, we first need to understand what this equation represents graphically. The absolute value equation |y| = f(x) is equivalent to two separate equations: y = f(x) and y = -f(x), provided f(x) ≥ 0.

In this case, the equation is |y| = 1 – x². This equation is valid only when $1 - x^2 \ge 0$, which means $x^2 \le 1$, or $-1 \le x \le 1$. For values of x in this range, the equation splits into two parts:

• y = 1 – x²
• y = -(1 – x²) = x² – 1

Let's analyze these two equations:

• The equation $y = 1 - x^2$ represents a parabola opening downwards, with its vertex at (0, 1). It intersects the x-axis when $1 - x^2 = 0$, which means $x = \pm 1$.
• The equation $y = x^2 - 1$ represents a parabola opening upwards, with its vertex at (0, -1). It intersects the x-axis when $x^2 - 1 = 0$, which means $x = \pm 1$.

The region bounded by |y| = 1 – x² for $-1 \le x \le 1$ is the region between the upper curve $y = 1 - x^2$ and the lower curve $y = x^2 - 1$ over the interval [-1, 1].

Calculating the Area Using Definite Integration

The area between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$, where $f(x) \ge g(x)$ over the interval [a, b], is given by the definite integral:

$\text{Area} = \int_{a}^{b} (f(x) - g(x)) dx$

In this problem:

• The upper curve is $f(x) = 1 - x^2$.
• The lower curve is $g(x) = x^2 - 1$.
• The interval is from $x = -1$ to $x = 1$.

So, the area is:

$\text{Area} = \int_{-1}^{1} ((1 - x^2) - (x^2 - 1)) dx$

Simplify the integrand:

$(1 - x^2) - (x^2 - 1) = 1 - x^2 - x^2 + 1 = 2 - 2x^2$

Now, evaluate the integral:

$\text{Area} = \int_{-1}^{1} (2 - 2x^2) dx$

Find the antiderivative of $2 - 2x^2$:

The antiderivative of 2 is $2x$.

The antiderivative of $-2x^2$ is $-2 \frac{x^{2+1}}{2+1} = -2 \frac{x^3}{3}$.

So, the antiderivative is $2x - \frac{2}{3}x^3$.

Evaluate the antiderivative at the limits of integration (1 and -1) and subtract:

$\text{Area} = [2x - \frac{2}{3}x^3]_{-1}^{1}$

$\text{Area} = (2(1) - \frac{2}{3}(1)^3) - (2(-1) - \frac{2}{3}(-1)^3)$

$\text{Area} = (2 - \frac{2}{3}) - (-2 - \frac{2}{3}(-1))$

$\text{Area} = (2 - \frac{2}{3}) - (-2 + \frac{2}{3})$

$\text{Area} = 2 - \frac{2}{3} + 2 - \frac{2}{3}$

$\text{Area} = 4 - \frac{4}{3}$

To subtract, find a common denominator:

$\text{Area} = \frac{4 \times 3}{3} - \frac{4}{3} = \frac{12}{3} - \frac{4}{3}$

$\text{Area} = \frac{12 - 4}{3} = \frac{8}{3}$

The area bounded by the curves |y| = 1 – x² is $\frac{8}{3}$ square units.

Revision Table: Key Concepts

Additional Information on Bounded Area Calculation

When dealing with areas bounded by curves involving absolute values, it's crucial to first understand how the absolute value affects the graph. The equation |y| = 1 – x² means that for any given x value (where $1 - x^2 \ge 0$), the y values are $\pm (1 - x^2)$. This creates symmetry about the x-axis.

Alternatively, because the region is symmetric about the x-axis and the y-axis, we could calculate the area in the first quadrant and multiply by 4. In the first quadrant, $x \ge 0$ and $y \ge 0$. The equation $y = 1 - x^2$ for $0 \le x \le 1$ bounds the area in the first quadrant above the x-axis. The area in the first quadrant bounded by $y = 1 - x^2$ and the x-axis from $x=0$ to $x=1$ is $\int_{0}^{1} (1 - x^2) dx$.

$\int_{0}^{1} (1 - x^2) dx = [x - \frac{x^3}{3}]_{0}^{1} = (1 - \frac{1^3}{3}) - (0 - \frac{0^3}{3}) = 1 - \frac{1}{3} = \frac{3-1}{3} = \frac{2}{3}$.

This $\frac{2}{3}$ represents the area in the first quadrant bounded by $y = 1-x^2$ and the x-axis. Similarly, in the fourth quadrant ($y \le 0$), the boundary is $y = -(1-x^2) = x^2 - 1$. The area between $y = x^2 - 1$ and the x-axis from $x=0$ to $x=1$ would be $\int_{0}^{1} (0 - (x^2 - 1)) dx = \int_{0}^{1} (1 - x^2) dx = \frac{2}{3}$ (taking the absolute value since area is positive). The total area is the sum of areas in all four quadrants, or because of symmetry, 4 times the area in the first quadrant bounded by $y = 1-x^2$ and $y=0$ (which is $x$-axis) for $x \in [0,1]$ is not correct, because the region $\vert y \vert = 1-x^2$ for $x \in [0,1]$ covers the area from $y=-(1-x^2)$ to $y=(1-x^2)$. The region is symmetric about both axes. We can calculate the area in the first quadrant, which is bounded by $y = 1-x^2$ and $x=0$ and $y=0$ up to $x=1$, plus the area in the fourth quadrant bounded by $y=x^2-1$ and $x=0$ and $y=0$ up to $x=1$. A simpler way is to use the symmetry about the y-axis and integrate from $x=0$ to $x=1$ the difference between the upper curve $y = 1-x^2$ and the lower curve $y = x^2-1$, and then multiply the result by 2. Area $= 2 \times \int_{0}^{1} ((1 - x^2) - (x^2 - 1)) dx = 2 \times \int_{0}^{1} (2 - 2x^2) dx$ Area $= 2 \times [2x - \frac{2}{3}x^3]_{0}^{1}$ Area $= 2 \times ((2(1) - \frac{2}{3}(1)^3) - (2(0) - \frac{2}{3}(0)^3))$ Area $= 2 \times ((2 - \frac{2}{3}) - (0 - 0))$ Area $= 2 \times (2 - \frac{2}{3}) = 2 \times (\frac{6 - 2}{3}) = 2 \times \frac{4}{3} = \frac{8}{3}$. This confirms the result obtained by integrating from -1 to 1.