surface de la partie bleu du graphe?


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Question Number 183426 by a.lgnaoui last updated on 25/Dec/22

$s u r f a c e d e l a p a r t i e b l e u$ $d u g r a p h e ?$
Commented by a.lgnaoui last updated on 25/Dec/22

Commented by cherokeesay last updated on 25/Dec/22

Commented by a.lgnaoui last updated on 26/Dec/22

$t h a n k y o u$
	Answered by Acem last updated on 26/Dec/22

	$S u r f . = \frac{1}{4} + \int_{1}^{\frac{3}{2}} (\frac{3}{2} x - x^{2}) d x$ $= \frac{1}{4} + {[x^{2} (\frac{3}{4} - \frac{1}{3} x)]}_{1}^{\frac{3}{2}}$ $= \frac{1}{4} + \frac{9}{16} - \frac{5}{12} = \frac{1}{4} + \frac{7}{48} = \frac{19}{48} {u n}^{2}$
	Commented by a.lgnaoui last updated on 26/Dec/22

	$t h a n k y o u$
	Answered by mr W last updated on 26/Dec/22

	$a w a y w i t h o u t i n t e g r a l :$ $x (x - 1) - 0.5 x = 0 \Rightarrow x^{2} - 1.5 x = 0$ $\Rightarrow A_{1} = \frac{{[{(- 1.5)}^{2} - 4 \times 1 \times 0]}^{3 / 2}}{6 \times 1^{2}} = \frac{27}{48}$ $x (x - 1) - 0 = 0 \Rightarrow x^{2} - x = 0$ $\Rightarrow A_{2} = \frac{{[{(- 1)}^{2} - 4 \times 1 \times 0]}^{3 / 2}}{6 \times 1^{2}} = \frac{1}{6}$ $A_{b l u e} = A_{1} - A_{2} = \frac{27}{48} - \frac{1}{6} = \frac{19}{48} ✓$
	Commented by Matica last updated on 27/Dec/22

	$h o w d i d y o u g e t t h i s f o r m u l a e ?$
	Commented by mr W last updated on 27/Dec/22

	$A = \frac{{(b^{2} - 4 a c)}^{3 / 2}}{6 a^{2}}$ $f o r m o r e s e e Q 89781$
	Commented by Matica last updated on 27/Dec/22

	$T h a n k y o u!$
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