Question

Let α be the area of the larger region bounded by the curve y² = 8x and the lines y = x and x = 2, which lies in the first quadrant. Then the value of 3α is equal to:

Answer 1

Correct Answer: 22

We are given the equations of two curves:

• \(y = x\)
• \(y^2 = 8x\)

Step 1: Points of Intersection

To find the points of intersection, solve \(y = x\) and \(y^2 = 8x\):

\[ x^2 = 8x \implies x(x - 8) = 0. \]

Thus, \(x = 0\) or \(x = 8\).

Substituting these values into \(y = x\), we get:

• When \(x = 0\), \(y = 0\).
• When \(x = 8\), \(y = 8\).

Thus, the points of intersection are \(O(0, 0)\) and \(A(8, 8)\).

The vertical line \(x = 2\) intersects the parabola \(y^2 = 8x\):

\[ y^2 = 16 \implies y = \pm 4. \]

This gives the points \(Q(2, 4)\) and \(R(2, -4)\).

Step 2: Area of the Shaded Region

The shaded region lies between the parabola \(y^2 = 8x\) and the line \(y = x\), from \(x = 2\) to \(x = 8\). The area is given by:

\[ \text{Area} = \int_{2}^{8} \left(\sqrt{8x} - x\right) dx. \]

Step 3: Solve Each Integral

First, split the integral:

\[ \text{Area} = \int_{2}^{8} \sqrt{8x} dx - \int_{2}^{8} x dx. \]

1. Solve \(\int_{2}^{8} \sqrt{8x} dx\):

Using substitution, let \(u = 8 - x\), so \(du = -dx\). The limits change as follows:

• When \(x = 2\), \(u = 6\).
• When \(x = 8\), \(u = 0\).

The integral becomes:

\[ \int_{2}^{8} \sqrt{8x} dx = 2\sqrt{2} \int_{6}^{0} u^{1/2} (-du). \]

Reversing the limits to eliminate the negative sign:

\[ \int_{2}^{8} \sqrt{8x} dx = 2\sqrt{2} \int_{0}^{6} u^{1/2} du. \]

Now integrate \(u^{1/2}\):

\[ \int u^{1/2} du = \frac{2}{3} u^{3/2}. \]

Apply the limits of integration:

\[ \int_{2}^{8} \sqrt{8x} dx = 2\sqrt{2} \cdot \frac{2}{3} \left[6^{3/2} - 0\right]. \]

Compute \(6^{3/2}\):

\[ 6^{3/2} = \sqrt{6^3} = \sqrt{216} = 6\sqrt{6}. \]

Thus:

\[ \int_{2}^{8} \sqrt{8x} dx = \frac{4\sqrt{2}}{3} \cdot 6\sqrt{6} = 4\sqrt{2} \cdot 2\sqrt{6} = 8\sqrt{12}. \]

2. Solve \(\int_{2}^{8} x dx\):

The integral is straightforward:

\[ \int x dx = \frac{x^2}{2}. \]

Apply the limits of integration:

\[ \int_{2}^{8} x dx = \frac{8^2}{2} - \frac{2^2}{2} = \frac{64}{2} - \frac{4}{2} = 32 - 2 = 30. \]

Step 4: Combine Results

The total area is:

\[ \text{Area} = \int_{2}^{8} \sqrt{8x} dx - \int_{2}^{8} x dx. \]

Substitute the results:

\[ \text{Area} = 8\sqrt{12} - 30. \]

Simplify further:

\[ \text{Area} = 22/3 \]