_0^2 x√(2-x) dx is equal to : Option 1: 8√2/15 Option 2: 2√2/15 Option 3: 16√2/15 Option 4: √2/15

Answer: 16√2/15. Solution: Put u = 2-x, so x = 2-u and the limits stay 0 to 2. The integral becomes _0^2 (2-u)√u du = [4/3u^(3/2)-2/5u^(5/2)]_0^2 = 8√2/3-8√2/5 = 16√2/15.

Calculus

Integrals

Medium

common

$\displaystyle\int_0^2 x\sqrt{2-x}\,dx$ is equal to :

$\dfrac{8\sqrt{2}}{15}$

$\dfrac{2\sqrt{2}}{15}$

$\dfrac{16\sqrt{2}}{15}$

$\dfrac{\sqrt{2}}{15}$

✅ Correct Option: 3

Put $u = 2-x$ , so $x = 2-u$ and the limits stay $0$ to $2$ . The integral becomes $\int_0^2 (2-u)\sqrt{u}\,du = \left[\frac{4}{3}u^{3/2}-\frac{2}{5}u^{5/2}\right]_0^2 = \frac{8\sqrt{2}}{3}-\frac{8\sqrt{2}}{5} = \frac{16\sqrt{2}}{15}$ .

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