Question

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Answer 1

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Answer 2

$\displaystyle I = \int_{0}^1x\sqrt{1-x^2}\, dx = -\dfrac{1}{2}\int_{0}^1(1-x^2)'\cdot (1-x^2)^{\frac{1}{2}}\, dx = \\ \\ = -\dfrac{1}{2}\cdot \dfrac{(1-x^2)^{\frac{3}{2}}}{\frac{3}{2}}\Bigg|_{0}^1 = -\dfrac{(1-x^2)^{\frac{3}{2}}}{3}\Bigg|_{0}^1 = 0+\dfrac{1}{3} = \boxed{\dfrac{1}{3}}$