Question

The region bounded by y=(3x)^(1/2), y=3x-6, y=0

Answer 1

Answer:

4.5 sq. units.

Step-by-step explanation:

The given curve is $y = (3x)^{\frac{1}{2} }$

⇒ $y^{2} = 3x$ ...... (1)

This curve passes through (0,0) point.

Now, the straight line is y = 3x - 6 ....... (2)

Now, solving (1) and (2) we get,

$y^{2} - y - 6 = 0$

⇒ (y - 3)(y + 2) = 0

⇒ y = 3 or y = -2

We will consider y = 3.

Now, y = 3x - 6 has zero at x = 2.

Therefor, the required are = $\int\limits^3_0 {(3x)^{\frac{1}{2} } } \, dx - \int\limits^3_2 {(3x - 6)} \, dx$

= $\sqrt{3} [{\frac{x^{\frac{3}{2} } }{\frac{3}{2} } }]^{3} _{0} - [\frac{3x^{2} }{2} - 6x ]^{3} _{2}$

= $[\frac{\sqrt{3}\times 2 \times 3^{\frac{3}{2} } }{3}] - [13.5 - 18 - 6 + 12]$

= 6 - 1.5

= 4.5 sq. units. (Answer)