Question-162651

Question Number 162651 by Mathematification last updated on 31/Dec/21

Answered by tounghoungko last updated on 31/Dec/21

y = - x {(x + a)}^{2} = - x^{3} - 2 {a x}^{2} - a^{2} x

y^{'} = - 3 x^{2} - 4 a x - a^{2} = 0

3 x^{2} + 4 a x + a^{2} = 0

(3 x + a) (x + a) = 0 {\begin{cases} x = - \frac{1}{3} a \\ x = - a \end{cases}

y^{″} = - 6 x - 4 a > 0 (f o r m i n i m u m)

x < - \frac{2}{3} a s o m i n a t x = - \frac{1}{3} a = \frac{2}{3}; a = - 2

∴ y = - x {(x - 2)}^{2}

i n t e r s e c t a t x - a x i s {\begin{cases} x = 0 \\ x = 2 \end{cases}

T h e y e l l o w a r e a i s

A = - \int_{0}^{2} - x {(x - 2)}^{2} d x

A = \int_{0}^{2} (x^{3} - 4 x^{2} + 4 x) d x

A = \frac{1}{4} . 16 - \frac{4}{3} . 8 + 2 . 2^{2} = \frac{4}{3}