Let-the-region-bounded-by-the-curve-y-x-1-x-and-the-x-axis-be-R-The-line-y-mx-divides-R-into-two-parts-find-the-value-of-1-m-3-

Question Number 161461 by ZiYangLee last updated on 18/Dec/21

Let the region bounded by the curve

y = x (1 - x) and the x - axis be R .

The line y = m x divides R into two parts,

find the value of {(1 - m)}^{3} .

Commented by cortano last updated on 18/Dec/21

T o t a l a r e a = \frac{2}{3} \times 1 \frac{1}{4} = \frac{1}{6}

A r e a I = \frac{Δ \sqrt{Δ}}{6 a^{2}}

w h e r e x^{2} + (m - 1) x = 0

Δ = {(m - 1)}^{2}; \sqrt{Δ} =∣ (m - 1) ∣

s i n c e m < 1

\Rightarrow a r e a I = \frac{1}{2} t o t a l a r e a

\Rightarrow \frac{{(1 - m)}^{3}}{6} = \frac{1}{12}

\Rightarrow {(1 - m)}^{3} = \frac{1}{2}

Commented by som(math1967) last updated on 18/Dec/21

I s t h e l i n e y = m x d i v i d e s R i n t o t w o e q u a l p a r t s

s i r ?

Commented by som(math1967) last updated on 18/Dec/21

x - x^{2} = 0

x (1 - x) = 0 ∴ x = 0, 1

a r e a o f R

\underset{0}{\int^{1}} (x - x^{2}) d x = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} s q u n i t

a g a i n m x = x - x^{2}

x = 0, (1 - m)

∴ \underset{0}{\int^{1 - m}} (x - x^{2}) d x - \underset{0}{\int^{1 - m}} m x d x = \frac{1}{12}

\frac{{(1 - m)}^{2}}{2} - \frac{{(1 - m)}^{3}}{3} - \frac{m {(1 - m)}^{2}}{2} = \frac{1}{12}

{(1 - m)}^{2} (\frac{1}{2} - \frac{1 - m}{3} - \frac{m}{2}) = \frac{1}{12}

{(1 - m)}^{2} (\frac{3 - 2 + 2 m - 3 m}{6}) = \frac{1}{12}

\frac{{(1 - m)}^{3}}{6} = \frac{1}{12}

∴ {(1 - m)}^{3} = \frac{1}{2}