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Question Number 192023 by Shlock last updated on 05/May/23

	Answered by a.lgnaoui last updated on 06/May/23
	$graphe(1) y _1 represente raphe is quart circle (radius 2) centre(0,0) x^2 +y^2 =4 0<x<2 y_1 =(√(4−x^2 )) (1) graphe(2) graphe is parabole: somet (0,2) coupe axe (o,x)en x=4 passant par point(5,−3) equation y_2 = −ax^2 +bx+c A(0,2) ⇒ c=2 B(4,0) et C(5;−3) ⇒ { ((−16a+4b+2=0)),((−25a+5b+2=−3)) :} 9a−b=3 ⇒ b=9a−3 −16a+4(9a−3)+2=0 ⇒(a=(1/2) b=(3/2)) y_2 =−(1/2)x^2 +(3/2)x+2 (2) Surface designee sur (fig 1) S=∫_0 ^4 (−(1/2)x^2 +(3/2)x+2)dx−∫_0 ^2 (√(4−x^2 )) dx [−(x^3 /6)+((3x^2 )/4)+2x+c]_(x=0) ^4 −2∣∫_0 ^2 (√(1−((x/2))^2 )) dx∣ S=9,34−4∫_0 ^(π/2) ∣ cos^2 tdt ∣ avec sin t=(x/2) dx=2cos tdt (a suivre).. .....$
	$graphe (1)$ $y_{1} represente raphe is quart circle (radius 2) centre (0, 0)$ $x^{2} + y^{2} = 4 0 < x < 2 y_{1} = \sqrt{4 - x^{2}} (1)$ $graphe (2)$ $graphe is p a rabole : somet (0, 2) coupe axe (o, x) en x = 4$ $passant par point (5, - 3)$ $equation y_{2} = - {ax}^{2} + bx + c$ $A (0, 2) \Rightarrow c = 2$ $B (4, 0) et C (5; - 3)$ $\Rightarrow {\begin{cases} - 16 a + 4 b + 2 = 0 \\ - 25 a + 5 b + 2 = - 3 \end{cases}$ $9 a - b = 3 \Rightarrow b = 9 a - 3$ $- 16 a + 4 (9 a - 3) + 2 = 0$ $\Rightarrow (a = \frac{1}{2} b = \frac{3}{2})$ $y_{2} = - \frac{1}{2} x^{2} + \frac{3}{2} x + 2 (2)$ $Surface designee sur (fig 1)$ $S = \int_{0}^{4} (- \frac{1}{2} x^{2} + \frac{3}{2} x + 2) dx - \int_{0}^{2} \sqrt{4 - x^{2}} dx$ ${[- \frac{x^{3}}{6} + \frac{{3 x}^{2}}{4} + 2 x + c]}_{x = 0}^{4} - 2 ∣ \int_{0}^{2} \sqrt{1 - {(\frac{x}{2})}^{2}} dx ∣$ $S = 9, 34 - 4 \int_{0}^{\frac{π}{2}} ∣ \cos^{2} tdt ∣$ $avec \sin t = \frac{x}{2} dx = 2 \cos tdt$ $(a suivre) . . . . . . .$
	Answered by a.lgnaoui last updated on 06/May/23
	$Surface pour graphe (fig 2) y_1 : A(2,2) B(3,5) C(7,7) forme d ′une parabole de somet C equation: −a_1 x^2 +b_1 x+c_1 x=2→ y=−4a_1 +2b_1 +c_1 =2 (1) x=3→ y=−9a_1 +3b_1 +c_1 =5 (2) x=7 → y=−49a_1 +7b_1 +c_1 =7 (3) (1)⇒ c_1 =2+4a_1 −2b_1 (2)−(3)→40a_1 −4b_1 =−2 ou 20a_1 −2b_1 =−1 (4) (2)→ −9a_1 +3b_1 +(2+4a_1 −2b_1 )=5 −5a_1 +b_1 =3 → { ((20a_1 −2b_1 =−1 )),((−5a_1 +b_1 =+3 )) :} 10a_1 =5 a_1 =(1/2) →b_1 =3+(5/2)=((11)/2) c_1 =2+2−11=7 determinant (((a_1 =−(1/2) b_1 =((11)/2) )),(( c_1 =−7))) donc y_1 → −(1/2)x^2 +((11)/2)x−7 Graphe(I) Calcul Equation (y_2 ) y_2 →A(0,0) ;B(4,2) C(5,5) y_2 : forme de parabole (centre (0,0) passant par A, B et C (croissante a_2 >0) y_2 =a_2 x^2 +b_2 x+c_2 x=0 y=0 ⇒ c_2 =0 x=4 16a_2 +4b_2 =2 ou 8a_2 +2b_2 =1 x=5 25a_2 +5b_2 =5 ou 5a_2 +b_2 =1 { ((4a_2 +2b_(2 ) = 1 (1))),((5a_2 + b_2 =1 (2))) :} ⇒a_2 =((1 )/6) b=(1/6) determinant (((a_2 =(1/6))),((b_2 =(1/6)))) y_2 =(x^2 /6)+(x/6) Surface sombre limite par A B C D S=∫_2 ^5 (y_1 −y_2 )dx−[(2×2)−∫_2 ^4 y_2 dx] −[(2×2)−∫_1 ^5 y_1 dx] ∫y_1 dx=((−x^3 )/6)+((11x^2 )/4)−7x ∫y_2 dx=(x^3 /(18))+(x^2 /(12)) S=[((−x^3 )/6)+((11x^2 )/4)−7x]_2 ^5 +[((−x^3 )/6)+((11x^2 )/4)−7x]_3 ^5 −[(x^3 /(18))+(x^2 /(12))]_2 ^5 +[(x^3 /(18))+(x^2 /(12))]_2 ^4 −y_1 (5)×(5−3)−4 apres des xalculs on obtient: Surface S=18−12 Soit S=6$
	$Surface pour graphe (fig 2)$ $y_{1} : A (2, 2) B (3, 5) C (7, 7)$ $forme d^{'} une parabole de somet C$ $equation : - a_{1} x^{2} + b_{1} x + c_{1}$ $x = 2 \to y = - 4 a_{1} + 2 b_{1} + c_{1} = 2 (1)$ $x = 3 \to y = - {9 a}_{1} + {3 b}_{1} + c_{1} = 5 (2)$ $x = 7 \to y = - {49 a}_{1} + {7 b}_{1} + c_{1} = 7 (3)$ $(1) \Rightarrow c_{1} = 2 + 4 a_{1} - 2 b_{1}$ $(2) - (3) \to {40 a}_{1} - {4 b}_{1} = - 2$ $ou 20 a_{1} - 2 b_{1} = - 1 (4)$ $(2) \to - {9 a}_{1} + {3 b}_{1} + (2 + {4 a}_{1} - {2 b}_{1}) = 5$ $- {5 a}_{1} + b_{1} = 3$ $\to {\begin{cases} {20 a}_{1} - {2 b}_{1} = - 1 \\ - {5 a}_{1} + b_{1} = + 3 \end{cases}$ $10 a_{1} = 5 a_{1} = \frac{1}{2} \to b_{1} = 3 + \frac{5}{2} = \frac{11}{2}$ $c_{1} = 2 + 2 - 11 = 7 \begin{array}{cc} a_{1} = - \frac{1}{2} b_{1} = \frac{11}{2} \\ c_{1} = - 7 \end{array}$ $donc$ $y_{1} \to - \frac{1}{2} x^{2} + \frac{11}{2} x - 7 Graphe (I)$ $Calcul Equat i on (y_{2})$ $y_{2} \to A (0, 0); B (4, 2) C (5, 5)$ $y_{2} : forme de parabole (centre (0, 0)$ $passant par A, B et C (croissante a_{2} > 0)$ $y_{2} = a_{2} x^{2} + b_{2} x + c_{2}$ $x = 0 y = 0 \Rightarrow c_{2} = 0$ $x = 4 16 a_{2} + 4 b_{2} = 2 ou$ $8 a_{2} + 2 b_{2} = 1$ $x = 5 25 a_{2} + 5 b_{2} = 5 ou$ $5 a_{2} + b_{2} = 1$ ${\begin{cases} 4 a_{2} + 2 b_{2} = 1 (1) \\ 5 a_{2} + b_{2} = 1 (2) \end{cases}$ $\Rightarrow a_{2} = \frac{1}{6} b = \frac{1}{6} \begin{array}{cc} a_{2} = \frac{1}{6} \\ b_{2} = \frac{1}{6} \end{array}$ $y_{2} = \frac{x^{2}}{6} + \frac{x}{6}$ $Surface sombre limite par A B C D$ $S = \int_{2}^{5} (y_{1} - y_{2}) dx - [(2 \times 2) - \int_{2}^{4} y_{2} dx]$ $- [(2 \times 2) - \int_{1}^{5} y_{1} dx]$ $\int y_{1} dx = \frac{- x^{3}}{6} + \frac{11 x^{2}}{4} - 7 x$ $\int y_{2} dx = \frac{x^{3}}{18} + \frac{x^{2}}{12}$ $S = {[\frac{- x^{3}}{6} + \frac{11 x^{2}}{4} - 7 x]}_{2}^{5} + {[\frac{- x^{3}}{6} + \frac{11 x^{2}}{4} - 7 x]}_{3}^{5}$ $- {[\frac{x^{3}}{18} + \frac{x^{2}}{12}]}_{2}^{5} + {[\frac{x^{3}}{18} + \frac{x^{2}}{12}]}_{2}^{4} - y_{1} (5) \times (5 - 3) - 4$ $apres des xalculs on obtient :$ $Surface S = 18 - 12$ $Soit S = 6$
	Commented by a.lgnaoui last updated on 06/May/23

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