Evaluate-5-5-25-x-2-dx-using-an-algebraic-method-Geometrical-mehod-thanks-in-advanced-great-mathematicians-

Question Number 72337 by Rio Michael last updated on 27/Oct/19

E v a l u a t e \int_{- 5}^{5} (\sqrt{25 - x^{2}}) d x u s i n g

\Rightarrow a n a l g e b r a i c m e t h o d

\Rightarrow G e o m e t r i c a l m e h o d

t h a n k s i n a d v a n c e d g r e a t m a t h e m a t i c i a n s

Commented by mathmax by abdo last updated on 27/Oct/19

a l g e b r i c m e t h o d l e t A = \int_{- 5}^{5} \sqrt{25 - x^{2}} d x \Rightarrow A = 2 \int_{0}^{5} \sqrt{25 - x^{2}} d x

(e v e n f u n c t i o n) \Rightarrow A =_{x = 5 s i n t} 2 \int_{0}^{\frac{π}{2}} \sqrt{25 - 25 {s i n}^{2} t} 5 c o s t d t

= 50 \int_{0}^{\frac{π}{2}} {c o s}^{2} t d t = 25 \int_{0}^{\frac{π}{2}} (1 + c o s (2 t)) d t

= \frac{25 π}{2} + \frac{25}{2} {[s i n (2 t)]}_{0}^{\frac{π}{2}} = \frac{25 π}{2}

Answered by MJS last updated on 27/Oct/19

geometrical method

y = \sqrt{25 - x^{2}} is the upper semicircle with center

(\begin{matrix} 0 \\ 0 \end{matrix}) and radius 5; integral = area \Rightarrow

\underset{- 5}{\int^{5}} \sqrt{25 - x^{2}} d x = \frac{5^{2} π}{2} = \frac{25}{2} π

algebraic method

\int \sqrt{25 - x^{2}} d x =

[t = \arcsin \frac{x}{5} \to d x = 5 \cos t d t]

= 25 \int \cos^{2} t d t = \frac{25}{2} \int 1 + \cos 2 t d t =

= \frac{25}{2} x + \frac{25}{4} \sin 2 t = \frac{25}{2} \arcsin \frac{x}{5} + \frac{1}{2} x \sqrt{25 - x^{2}} + C

now use borders

Commented by Rio Michael last updated on 27/Oct/19

t h a n k s s i r