Question

The area of the region enclosed by the curves \( y = x^2 - 4x + 4 \) and \( y^2 = 16 - 8x \) is:

• A. \( \frac{4}{3} \)
• B. \( 8 \)
• C. \( \frac{8}{3} \)
• D. \( 5 \)

Hint:

When finding the area between curves: - Set up the appropriate integral by first determining the points of intersection. - Integrate the difference between the two curves over the appropriate interval. - Always check the limits of integration and the nature of the curves involved.

Answer 1

The Correct Option is C

To determine the area bounded by the specified curves, the initial step involves identifying their points of intersection. The curves provided are \( y = x^2 - 4x + 4 \) and \( y^2 = 16 - 8x \).

1. Determine Intersection Points:
From the equation \( y^2 = 16 - 8x \), we derive \( y = \pm\sqrt{16 - 8x} \).
Substituting \( y = x^2 - 4x + 4 \) into \( y^2 = 16 - 8x \) yields:

\((x^2 - 4x + 4)^2 = 16 - 8x\)

Upon expansion and simplification, we get:
\((x^2 - 4x + 4)^2 = x^4 - 8x^3 + 24x^2 - 32x + 16\)
Thus, the equation becomes: \(x^4 - 8x^3 + 24x^2 - 32x + 16 = 16 - 8x\)
Rearranging the terms gives:
\(x^4 - 8x^3 + 24x^2 - 24x = 0\)
Factoring the equation produces:
\(x(x^3 - 8x^2 + 24x - 24) = 0\)

Using trial division with \( (x - 2) \) as a potential factor:
Performing synthetic or polynomial division results in:
\((x - 2)(x^2 - 6x + 12)\)

Solving the equation \(x(x - 2)(x^2 - 6x + 12) = 0\), the real roots are \(x = 0\) and \(x = 2\). The quadratic \(x^2 - 6x + 12\) yields no real roots.
The intersection points are thus at \(x = 0\) and \(x = 2\).

2. Calculate the Area:
The upper boundary curve is \(y = \sqrt{16 - 8x}\) and the lower boundary curve is \(y = x^2 - 4x + 4\).

The area \(A\) is computed using the integral:

\[A = \int_{0}^{2} [\sqrt{16 - 8x} - (x^2 - 4x + 4)] \, dx\]

The integral can be separated as follows:
\(A = \int_{0}^{2} \sqrt{16 - 8x} \, dx - \int_{0}^{2} (x^2 - 4x + 4) \, dx\)

Now, we evaluate each integral:

Integral 1:
\(\int \sqrt{16 - 8x} \, dx\)
Employing the substitution \(u = 16 - 8x\), so \(du = -8 \, dx\), which implies \(dx = -\frac{1}{8} \, du\):
\(\int \sqrt{u} \, \left( -\frac{1}{8} \right) \, du = -\frac{1}{8} \cdot \frac{2}{3} u^{3/2} = -\frac{1}{12} \cdot (16 - 8x)^{3/2} \)

Evaluating this from 0 to 2:
\(-\frac{1}{12}[(16 - 8(2))^{3/2} - (16 - 8(0))^{3/2}]\) is incorrect. The evaluation should be: \(\left[-\frac{1}{12} (16 - 8x)^{3/2}\right]_{0}^{2} = -\frac{1}{12}[(16 - 16)^{3/2} - (16 - 0)^{3/2}] = -\frac{1}{12}[0 - 64] = \frac{64}{12}\)

Integral 2:
\(\int_{0}^{2} (x^2 - 4x + 4) \, dx = \left[\frac{x^3}{3} - 2x^2 + 4x \right]_{0}^{2}\)
The definite integral yields:
\(\left(\frac{2^3}{3} - 2(2)^2 + 4(2)\right) - \left(\frac{0^3}{3} - 2(0)^2 + 4(0)\right) = \left[\frac{8}{3} - 8 + 8 \right] - 0 = \frac{8}{3}\)

Final Area Calculation:
\(A = \frac{64}{12} - \frac{8}{3}\)
Simplifying, \(A = \frac{16}{3} - \frac{8}{3} = \frac{8}{3}\)

Consequently, the area of the enclosed region is \(\frac{8}{3}\).