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Question Number 189825 by TUN last updated on 22/Mar/23

{\int_{a}}^{b} \sqrt{(x - a) (b - x)} = ¿

Answered by Frix last updated on 14/Apr/23

circle :

{(x - \frac{a + b}{2})}^{2} + y^{2} = \frac{{(a - b)}^{2}}{4} = r^{2} \Rightarrow r = \frac{∣ a - b ∣}{2}

\Leftrightarrow

y = \pm \sqrt{(x - a) (b - x)}

\Rightarrow

\int \sqrt{(x - a) (b - x)} d x = \frac{r^{2} π}{2} = \frac{{(a - b)}^{2} π}{8}

Answered by Mathspace last updated on 24/Mar/23

\sqrt{x - a} = t \Rightarrow x - a = t^{2} \Rightarrow x = t^{2} + a

I = \int_{0}^{\sqrt{b - a}} t \sqrt{b - a - t^{2}} (2 t) d t

= 2 \int_{0}^{\sqrt{b - a}} t^{2} \sqrt{b - a - t^{2}} d t

=_{t = \sqrt{b - a} y} 2 \int_{0}^{1} (b - a) y^{2} \sqrt{b - a} \sqrt{1 - y^{2}} \sqrt{b - a} d y

= 2 {(b - a)}^{2} \int_{0}^{1} y^{2} \sqrt{1 - y^{2}} d y

\int_{0}^{1} y^{2} \sqrt{1 - y^{2}} d y =_{y = s i n t} \int_{0}^{\frac{π}{2}} {s i n}^{2} t c o s t c o s t d t

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

= \frac{1}{4} \int_{0}^{\frac{π}{2}} {s i n}^{2} (2 t) d t

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

= \frac{1}{8} . \frac{π}{2} - \frac{1}{8.4} {[s i n (4 t)]}_{0}^{\frac{π}{2}} (\to 0)

= \frac{π}{16} \Rightarrow I = 2 {(b - a)}^{2} . \frac{π}{16}

= \frac{π}{8} {(b - a)}^{2}

h e r e w e h a v e s u p p o s e d a ⩽ b