By using the properties of definite integrals,evaluate the integral$\int\limits_0^2x\sqrt{2-x}dx$

$\begin{array}{1 1} \frac{16 \sqrt 2 } {15} \\ \frac{32 \sqrt 2}{15} \\ \frac{30 \sqrt 8}{32} \\ \frac{16 \sqrt 4}{15} \end{array}$

Toolbox:
• (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
• (ii)$\int \limits_0^af(x)dx=\int \limits_0^af(a-x)dx$
Given $I=\int\limits_0^2x\sqrt{2-x}dx$

Applying the properties:$\int \limits_0^af(x)dx=\int \limits_0^af(a-x)dx$

$I=\int\limits_0^2(2-x)\sqrt{2-2+x}dx$

$I=\int\limits_0^2(2-x)\sqrt{x}dx$

On seperating the term

$=\int\limits_0^2 2\sqrt{x}-\int\limits_0^2 x\sqrt{x}dx$

$=\int\limits_0^2 2(x)^{1/2}-\int\limits_0^2 x^{3/2}dx$

On integrating we get,

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

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

On applying limits

$=\frac{4}{3}\bigg[2^{3/2}-0\bigg]-\frac{2}{5}\bigg[2^{5/2}-0\bigg]$

$=\frac{4}{3} \times 2\sqrt 2-\frac{2}{5} \times 4 \sqrt 2$

$= \frac{8\sqrt 2}{3}-\frac{\sqrt 2}{5}=\frac{8 \times 5 \sqrt 2-8 \times 3\sqrt 2}{15}$

$=\frac{40\sqrt 2-24 \sqrt 2}{15}=\frac{16 \sqrt 2}{15}$

Hence $\int \limits_0^2 x \sqrt {2-x}dx=\large\frac{16 \sqrt 2}{15}$