calculate-0-1-x-1-x-dx-

Question Number 204902 by pticantor last updated on 01/Mar/24

c a l c u l a t e \int_{0}^{1} \sqrt{x (1 - x)} d x

Answered by witcher3 last updated on 01/Mar/24

y^{2} + x^{2} - x = 0 \Leftrightarrow y^{2} + {(x - \frac{1}{2})}^{2} = \frac{1}{4}

circle center (\frac{1}{2}, 0); R = \frac{1}{2}

S = π . \frac{1}{4};

our region is ? y ⩾ 0

\frac{S}{2} = \frac{π}{8}

Commented by pticantor last updated on 01/Mar/24

g r e a t

t h k s t o o m u c h

Answered by Mathspace last updated on 01/Mar/24

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

x = {s i n}^{2} t \Rightarrow I = \int_{0}^{\frac{π}{2}} \sqrt{{s i n}^{2} t . {c o s}^{2} t} 2 s i n t c o s t d t

= 2 \int_{0}^{\frac{π}{2}} {s i n}^{2} t \times {c o s}^{2} t d t

= 2 \int_{0}^{\frac{π}{2}} {(\frac{1}{2} s i n (2 t))}^{2} d t

= \frac{1}{2} \int_{0}^{\frac{π}{2}} {s i n}^{2} (2 t) d t

= \frac{1}{4} \int_{0}^{\frac{π}{2}} (1 - c o s (4 t)) d t

= \frac{π}{8} - \frac{1}{16} {[s i n (4 t)]}_{0}^{\frac{π}{2}}

= \frac{π}{8} - 0 = \frac{π}{8}