Question

Area lying between the curves \(y^2 = 9x\) and y = 3x is:

• A. \(\frac {1}{4} sq. units\)
• B. \(\frac {1}{2} sq. units\)
• C. \(\frac {2}{3} sq. units\)
• D. \(\frac {1}{3} sq. units\)

Answer 1

The Correct Option is B

The problem requires finding the area between the curves represented by the equations \(y^2 = 9x\) and \(y = 3x\).

First, let's find the points of intersection for these curves by equating the given equations:

• From \(y = 3x\), substitute into \(y^2 = 9x\):
• \((3x)^2 = 9x\)
• \(9x^2 = 9x\)
• Divide through by 9:
• \(x^2 = x\)
• \(x^2 - x = 0\)
• \(x(x-1) = 0\)
• Thus, \(x = 0\) or \(x = 1\).

These correspond to:

• For \(x=0\): \(y = 3 \times 0 = 0\), giving point \((0,0)\).
• For \(x=1\): \(y = 3 \times 1 = 3\), giving point \((1,3)\).

The area of interest is bounded between \(x = 0\) and \(x = 1\). The equations are \(y = \sqrt{9x}\) and \(y = 3x\). We calculate the area between these curves by integrating:

\(\text{Area} = \int_0^1 (\sqrt{9x} - 3x) \, dx\)

Break down the integral as follows:

• \(\int \sqrt{9x} \, dx = \int 3\sqrt{x} \, dx = 3 \cdot \frac{2}{3}x^{3/2} = 2x^{3/2}\)
• \(\int 3x \, dx = \frac{3}{2}x^2\)

Evaluate both from 0 to 1:

• \([2x^{3/2}]_0^1 = [2(1)^{3/2} - 2(0)^{3/2}] = 2\)
• \([\frac{3}{2}x^2]_0^1 = [\frac{3}{2}(1)^2 - \frac{3}{2}(0)^2] = \frac{3}{2}\)

Subtract the two results to find the area:

\(\text{Area} = 2 - \frac{3}{2} = \frac{1}{2} \, \text{sq. units}\)