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.

Answer 2

We want to find the area of the region bounded by the curve $ y^2 = 8x $ and the line $ y = 2x $.

First, find the intersection points:

$$ (2x)^2 = 8x \implies 4x^2 = 8x \implies 4x^2 - 8x = 0 \implies 4x(x-2) = 0 \implies x=0 \text{ or } x=2. $$

When $ x=0 $, $ y=2(0) = 0 $. When $ x=2 $, $ y=2(2) = 4 $. The intersection points are $ (0,0) $ and $ (2,4) $.

The area is given by:

$$ \int_0^2 (\sqrt{8x} - 2x) dx. $$

Split the integral:

$$ \int_0^2 (\sqrt{8x} - 2x) dx = \sqrt{8} \int_0^2 x^{1/2} dx - 2 \int_0^2 x dx. $$

Evaluate each term separately. For the first term:

$$ \sqrt{8} \int_0^2 x^{1/2} dx = 2\sqrt{2} \cdot \frac{x^{3/2}}{3/2} \Big|_0^2 = 2\sqrt{2} \cdot \frac{2}{3} \cdot (2^{3/2} - 0) = 2\sqrt{2} \cdot \frac{2}{3} \cdot 2\sqrt{2} = \frac{4\sqrt{2}}{3} \cdot 2\sqrt{2} = \frac{16}{3}. $$

For the second term:

$$ 2 \int_0^2 x dx = 2 \cdot \frac{x^2}{2} \Big|_0^2 = (2^2 - 0) = 4. $$

Combine the results:

$$ \int_0^2 (\sqrt{8x} - 2x) dx = \frac{16}{3} - 4 = \frac{16}{3} - \frac{12}{3} = \frac{4}{3}. $$

The area of the region bounded by the curve and the line is $ \frac{4}{3} $ square units.