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

Answer:4.5 sq. units.Step-by-step explanation:The given curve is [tex]y = (3x)^{\frac{1}{2} }[/tex]⇒ [tex]y^{2} = 3x[/tex] ...... (1)This curve passes through (0,0) point.Now, the straight line is y = 3x - 6 ....... (2)Now, solving (1) and (2) we get, [tex]y^{2} - y - 6 = 0[/tex]⇒ (y - 3)(y + 2) = 0⇒ y = 3 or y = -2We will consider y = 3.Now, y = 3x - 6 has zero at x = 2.Therefor, the required are = [tex]\int\limits^3_0 {(3x)^{\frac{1}{2} } } \, dx - \int\limits^3_2 {(3x - 6)} \, dx[/tex]= [tex]\sqrt{3} [{\frac{x^{\frac{3}{2} } }{\frac{3}{2} } }]^{3} _{0} - [\frac{3x^{2} }{2} - 6x ]^{3} _{2}[/tex]= [tex][\frac{\sqrt{3}\times 2 \times 3^{\frac{3}{2} }  }{3}] - [13.5 - 18 - 6 + 12][/tex]= 6 - 1.5 = 4.5 sq. units. (Answer)