find-a-b-b-x-x-a-dx-with-a-lt-b-then-find-1-2-2-x-x-1-dx-

Question Number 31074 by abdo imad last updated on 02/Mar/18

f i n d \int_{a}^{b} \sqrt{(b - x) (x - a)} d x w i t h a < b . t h e n f i n d

\int_{1}^{\sqrt{2}} \sqrt{(\sqrt{2} - x) (x - 1)} d x .

Commented by abdo imad last updated on 09/Mar/18

l e t p u t I = \int_{a}^{b} \sqrt{(b - x) (x - a)} d x t h e c h . x = \frac{a - b}{2} t + \frac{a + b}{2}

b - x = b - \frac{a + b}{2} - \frac{a - b}{2} t = \frac{b - a}{2} - \frac{a - b}{2} t = \frac{b - a}{2} (1 + t)

x - a = \frac{a - b}{2} t + \frac{a + b}{2} - a = \frac{a - b}{2} t + \frac{b - a}{2} = \frac{b - a}{2} (1 - t) a n d

w e h a v e \frac{a - b}{2} t = x - \frac{a + b}{2} \Rightarrow t = \frac{2 (x - \frac{a + b}{2})}{a - b} = \frac{2 x - a - b}{a - b}

x = a \Rightarrow t = 1 a n d x = b \Rightarrow t = - 1 \Rightarrow

I = - \int_{- 1}^{1} \sqrt{\frac{b - a}{2} (1 + t) \frac{b - a}{2} (1 - t)} \frac{a - b}{2} d t

I = \frac{{(b - a)}^{2}}{4} \int_{- 1}^{1} \sqrt{1 - t^{2}} d t = \frac{{(b - a)}^{2}}{2} \int_{0}^{1} \sqrt{1 - t^{2}} d t c h . t = s i n θ

g i v e \int_{0}^{1} \sqrt{1 - t^{2}} d t = \int_{0}^{\frac{π}{2}} c o s θ c o s θ d θ = \int_{0}^{\frac{π}{2}} {c o s}^{2} θ d θ

= \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 + c o s (2 θ) d θ = \frac{π}{4} \Rightarrow I = \frac{{(b - a)}^{2}}{8} . t h e n l e t t a k e

a = 1 a n d b = \sqrt{2} w e g e t

\int_{1}^{\sqrt{2}} \sqrt{(\sqrt{2} - x) (x - 1)} d x = \frac{{(\sqrt{2} - 1)}^{2}}{8} = \frac{3 - 2 \sqrt{2}}{8} .