n-0-1-x-R-f-x-n-n-1-x-n-The-given-function-gives-a-linear-line-that-goes-through-points-0-n-and-1-n-0-The-function-changes-as-n-changes-What-area-is-beneath-the-shape-made-as-n-goe

Question Number 106867 by Penguin last updated on 07/Aug/20

\forall n \in (0, 1) \forall x \in R : f (x) = \frac{n}{n - 1} x + n

The given function gives a linear line that

goes through points (0, n) and (1 - n, 0) .

The function changes as n changes .

What area is beneath the shape made as

n goes from 0 \to 1 ?

Commented by Penguin last updated on 07/Aug/20

Is the solution :

A = \int_{0}^{1} \int_{0}^{1} f d x d n ? ?

Commented by Penguin last updated on 07/Aug/20

Commented by Penguin last updated on 07/Aug/20

Always makes triangle with A = \frac{1}{2} n (1 - n)

A = \frac{1}{n} (t_{1} + t_{2} + \dots t_{n})

t_{n} = sum of n th triangle

t_{n} = \frac{1}{2} n (1 - n)

t_{n + 1} \frac{1}{2} (n + ε) (1 - n - ε)

for infinit triangles :

A = \frac{1}{1} \int_{0}^{1} \frac{1}{2} n (1 - n) d n

A = \frac{1}{2} {[\frac{1}{2} n^{2} - \frac{1}{3} n^{3}]}_{0}^{1}

A = \frac{1}{12}

Thks looks wrong, as the plot looks like :

A = 1 - \frac{π}{4} (\frac{1}{4} circle in a sqare)

Answered by Her_Majesty last updated on 07/Aug/20

f_{p} : y = \frac{p}{p - 1} x + p

f_{q} : y = \frac{q}{q - 1} x + q

f_{p} \cap f_{q} :

\frac{p}{p - 1} x + p = \frac{q}{q - 1} x + q

x = (p - 1) (q - 1)

y = p q

n o w q \to p

x = {(p - 1)}^{2}

y = p^{2}

p = 1 \pm \sqrt{x}

y = {(1 \pm \sqrt{x})}^{2} = 1 \pm 2 \sqrt{x} + x

y o u r c u r v e y = 1 - 2 \sqrt{x} + x

\underset{0}{\int^{1}} (1 - 2 \sqrt{x} + x) d x = {[x - \frac{4}{3} x^{3 / 2} + \frac{1}{2} x^{2}]}_{0}^{1} = \frac{1}{6}

n o t s u r e i f t h i s m a k e s s e n s e \dots

Commented by mr W last updated on 07/Aug/20

a n s w e r i s c o r r e c t!

Answered by mr W last updated on 07/Aug/20

\frac{y}{n} + \frac{x}{1 - n} = 1 \dots (i) e q n . o f L_{1}

\frac{y}{n + d n} + \frac{x}{1 - n - d n} = 1 \dots (i i) e q n . o f L_{2}

i n t e r s e c t i o n o f (i) a n d (i i) a t (x_{1}, y_{1})

\Rightarrow \frac{y_{1}}{n (1 - n - d n)} - \frac{y_{1}}{(n + d n) (1 - n)} = \frac{1}{1 - n - d n} - \frac{1}{1 - n}

\Rightarrow y_{1} = n (n + d n)

d A = \frac{y_{1} d n}{2} = \frac{n (n + d n) d n}{2} = \frac{n^{2} d n}{2}

A = \frac{1}{2} \int_{0}^{1} n^{2} d n = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}

Commented by Her_Majesty last updated on 07/Aug/20

t h a n k y o u

Commented by mr W last updated on 07/Aug/20

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