Question

$\int\limits^3_0x²(3-x)¹/² dx=144√3/35 ,Prove it.$

Answer 1

Answer:

$\int\limits^3_0\:x^2(3-x)^{\frac{1}{2}}\:dx=\frac{144\sqrt3}{35}$

Step-by-step explanation:

$Let\:I=\int\limits^3_0\:x^2(3-x)^{\frac{1}{2}}\:dx$

using the property

$\boxed{\int\limits^a_0\:f(x)\:dx=\int\limits^a_0\:f(a-x)\:dx}$

$\implies\:I=\int\limits^3_0\:(3-x)^2(3-(3-x))^{\frac{1}{2}}\:dx$

$\implies\:I=\int\limits^3_0\:[9+x^2-6x]x^{\frac{1}{2}}\:dx$

$\implies\:I=\int\limits^3_0\:[9x^{\frac{1}{2}}+x^{\frac{5}{2}}-6x^{\frac{3}{2}}]\:dx$

$\implies\:I=[9\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{7}{2}}}{\frac{7}{2}}-6\frac{x^{\frac{5}{2}}}{\frac{5}{2}}]\limits^3_0$

$\implies\:I=2[9\frac{x^{\frac{3}{2}}}{3}+\frac{x^{\frac{7}{2}}}{7}-6\frac{x^{\frac{5}{2}}}{5}]\limits^3_0$

$\implies\:I=2[3(3^{\frac{3}{2}})+\frac{3^{\frac{7}{2}}}{7}-6\frac{3^{\frac{5}{2}}}{5}]-2[0]$

$\implies\:I=2[3(3\sqrt3)+\frac{3^3\sqrt3}{7}-6\frac{3^2\sqrt3}{5}]$

$\implies\:I=2(9)[\sqrt3+\frac{3\sqrt3}{7}-6\frac{\sqrt3}{5}]$

$\implies\:I=18[\frac{35\sqrt3+15\sqrt3-42\sqrt3}{35}]$

$\implies\:I=18[\frac{8\sqrt3}{35}]$

$\therefore\boxed{\bf\:I=\frac{144\sqrt3}{35}}$