find-0-2-x-3-2-x-dx-

Question Number 66340 by mathmax by abdo last updated on 12/Aug/19

f i n d \int_{0}^{2} \sqrt{x^{3} (2 - x)} d x

Commented by mathmax by abdo last updated on 15/Aug/19

l e t I = \int_{0}^{2} \sqrt{x^{3} (2 - x)} d x \Rightarrow I = \int_{0}^{2} x \sqrt{x (2 - x)} d x

= \int_{0}^{2} x \sqrt{- x^{2} + 2 x} d x w e h a v e - x^{2} + 2 x = - (x^{2} - 2 x)

= - (x^{2} - 2 x + 1 - 1) = - {(x - 1)}^{2} + 1 \Rightarrow I = \int_{0}^{2} x \sqrt{1 - {(x - 1)}^{2}} d x

w e u s e t h e c h a n g e m e n t x - 1 = s i n t \Rightarrow

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} (1 + s i n t) c o s t (c o s t) d t = \int_{- \frac{π}{2}}^{\frac{π}{2}} {c o s}^{2} t d t + \int_{- \frac{π}{2}}^{\frac{π}{2}} s i n t {c o s}^{2} t d t

= \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + c o s (2 t)}{2} d t + 0 (t \to s i n t {c o s}^{2} t i s o d d)

= \frac{π}{2} + \frac{1}{4} {[s i n (2 t)]}_{- \frac{π}{2}}^{\frac{π}{2}} = \frac{π}{2} + 0 \Rightarrow I = \frac{π}{2}