Question

Mă poate ajuta cineva va rog mult?​

Answer 1

Răspuns:

$\int_{0}^{1}{x\sqrt{1-x^2}dx}=\frac{1}{3}$

Explicație pas cu pas:

$\int_{0}^{1}{x\sqrt{1-x^2}dx}\\</p><p>=\dfrac{-1}{2}\int_{0}^{1}{-2x\sqrt{1-x^2}dx}\\</p><p>=-\dfrac{1}{2}\int_{0}^{1}{\sqrt{1-x^2}d\left(1-x^2\right)}\\</p><p>=-\dfrac{1}{2}\left[\dfrac{(1-x^2)^{3/2}}{3/2}\right]_{0}^{1}\\</p><p>=-\dfrac{1}{2}\left(-\dfrac{2}{3}\right)\\</p><p>=\dfrac{1}{3}$

$\hfill{\boxdot}$