surface-de-la-partie-bleu-du-graphe-

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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