Question

By using the properties of definite integrals,evaluate the integrals: $\int^2_0 {x \sqrt{2-x}} \, dx$

Answer 1

Answer:

Step-by-step explanation:

Formula used:

$\int{x^n}\:dx=\frac{x^{n+1}}{n+1}+c$

Now,

$\int\limits^2_0{x\sqrt{2-x}}\:dx\\\\=\int\limits^2_0{[2-(2-x)]\sqrt{2-x}}\:dx\\\\=\int\limits^2_0{[2-(2-x)](2-x)^{\frac{1}{2}}}\:dx$

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

$=[2\frac{(2-x)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(2-x)^{\frac{5}{2}}}{\frac{5}{2}}]^2_0$

$=[2\frac{(2-2)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(2-2)^{\frac{5}{2}}}{\frac{5}{2}}]-[2\frac{(2-0)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(2-0)^{\frac{5}{2}}}{\frac{5}{2}}]$

$=-[\frac{2}{3}2^{\frac{5}{2}}-\frac{2}{5}2^{\frac{5}{2}}]$

$=-[\frac{2}{3}-\frac{2}{5}]2^{\frac{5}{2}}$

$=-[\frac{10-6}{15}]2^{\frac{5}{2}}$

$=-[\frac{4}{15}]2^{\frac{5}{2}}$