Question

The area of the region bounded by the curve y²=8x and the line y = 2x is

• A. \(\frac{16}{3}\)sq. units
• B. \(\frac{4}{3}\)sq. units
• C. \(\frac{3}{4}\)sq. units
• D. \(\frac{8}{3}\)sq. units

Answer 1

The Correct Option is B

We need to find the area of the region bounded by the curve \(y^2 = 8x\) and the line \(y = 2x\).

First, find the points of intersection by setting the equations equal to each other:

\(y^2 = 8x\) and \(y = 2x\). Substitute \(y = 2x\) into the first equation: \((2x)^2 = 8x\)

\(4x^2 = 8x\)

\(4x^2 - 8x = 0\)

\(4x(x - 2) = 0\)

\(x = 0\) or \(x = 2\)

When \(x = 0\), \(y = 2(0) = 0\).

When \(x = 2\), \(y = 2(2) = 4\).

So, the points of intersection are (0, 0) and (2, 4).

Now we set up the integral for the area between the curves. Since \(y^2 = 8x\), \(x = \frac{y^2}{8}\).

We want to integrate with respect to y, so we express x in terms of y. The line is \(x = \frac{y}{2}\).

We integrate from y = 0 to y = 4. The area is given by:

\(\int_{0}^{4} (\frac{y}{2} - \frac{y^2}{8}) \, dy\)

\(= [\frac{y^2}{4} - \frac{y^3}{24}]_{0}^{4}\)

\(= (\frac{4^2}{4} - \frac{4^3}{24}) - (0 - 0)\)

\(= (\frac{16}{4} - \frac{64}{24})\)

\(= 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}\)

Therefore, the area of the region is \(\frac{4}{3}\) square units.

Thus, the correct option is (B) \(\frac{4}{3}\) sq. units.