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 find the area enclosed by the given curves, we first need to identify the points of intersection. The curves are \( y = x^2 - 4x + 4 \) and \( y^2 = 16 - 8x \).

1. Solve for Points of Intersection:
From \( y^2 = 16 - 8x \), we have \( y = \pm\sqrt{16 - 8x} \).
Substitute \( y = x^2 - 4x + 4 \) into \( y^2 = 16 - 8x \):

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

Expand and simplify:
\((x^2 - 4x + 4)^2 = x^4 - 8x^3 + 24x^2 - 32x + 16\)
So: \(x^4 - 8x^3 + 24x^2 - 32x + 16 = 16 - 8x\)
Rearrange terms:
\(x^4 - 8x^3 + 24x^2 - 24x = 0\)
Factor the equation:
\(x(x^3 - 8x^2 + 24x - 24) = 0\)

Find roots using trial for \((x - 2)\) as a factor:
Perform synthetic division or polynomial division to get:
\((x - 2)(x^2 - 6x + 12)\)

Solve \(x(x - 2)(x^2 - 6x + 12) = 0\)
Roots are \(x = 0\) and \(x = 2\). The quadratic \(x^2 - 6x + 12\) has no real roots.
Intersection Points: \(x = 0, 2\).

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

Area \(A\) is given by:

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

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

Now solve each integral:

Integral 1:
\(\int \sqrt{16 - 8x} \, dx\)
Substitute \(u = 16 - 8x\), \(du = -8 \, dx\). Hence, \(dx = -\frac{1}{8} \, du\):
\(\int \sqrt{u} \, \left( -\frac{1}{8} \right) \, du = -\frac{1}{8} \cdot \frac{2}{3} u^{3/2} = -\frac{1}{12} \cdot (16 - 8x)^{3/2} \)

Evaluate from 0 to 2:
\(-\frac{1}{12}[(16 - 0)^{3/2} - (16 - 16)^{3/2}]\)
\(-\frac{1}{12}[64 - 0] = \frac{64}{12}\)

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

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

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

Answer 2

Given curves: \[ y = (x - 2)^2, \quad y^2 = 8(x - 2) \] The equation of the second curve is: \[ y = x^2, \quad y^2 = -8x \] Now, to find the area of the region enclosed between these curves, we calculate as follows: \[ \text{Area} = \frac{16ab}{3} = \frac{16 \times \frac{1}{4} \times 2}{3} = \frac{8}{3} \] Thus, the area between the curves is \({\frac{8}{3}} \).