∀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 goes from 0→1?


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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

$Always makes triangle with A = \frac{1}{2} n (1 - n)$ $A = \frac{1}{n} (t_{1} + t_{2} + . . . t_{n})$ $t_{n} = sum of n th triangle$ $t_{n} = \frac{1}{2} n (1 - n)$ $t_{n + 1} \frac{1}{2} (n + ε) (1 - n - ε)$ $for infinit triangles :$ $A = \frac{1}{1} \int_{0}^{1} \frac{1}{2} n (1 - n) d n$ $A = \frac{1}{2} {[\frac{1}{2} n^{2} - \frac{1}{3} n^{3}]}_{0}^{1}$ $A = \frac{1}{12}$ $Thks looks wrong, as the plot looks like :$ $A = 1 - \frac{π}{4} (\frac{1}{4} circle in a sqare)$
	Answered by Her_Majesty last updated on 07/Aug/20

	$f_{p} : y = \frac{p}{p - 1} x + p$ $f_{q} : y = \frac{q}{q - 1} x + q$ $f_{p} \cap f_{q} :$ $\frac{p}{p - 1} x + p = \frac{q}{q - 1} x + q$ $x = (p - 1) (q - 1)$ $y = p q$ $n o w q \to p$ $x = {(p - 1)}^{2}$ $y = p^{2}$ $p = 1 \pm \sqrt{x}$ $y = {(1 \pm \sqrt{x})}^{2} = 1 \pm 2 \sqrt{x} + x$ $y o u r c u r v e y = 1 - 2 \sqrt{x} + x$ $\underset{0}{\int^{1}} (1 - 2 \sqrt{x} + x) d x = {[x - \frac{4}{3} x^{3 / 2} + \frac{1}{2} x^{2}]}_{0}^{1} = \frac{1}{6}$ $n o t s u r e i f t h i s m a k e s s e n s e . . .$
	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

	$\frac{y}{n} + \frac{x}{1 - n} = 1 . . . (i) e q n . o f L_{1}$ $\frac{y}{n + d n} + \frac{x}{1 - n - d n} = 1 . . . (i i) e q n . o f L_{2}$ $i n t e r s e c t i o n o f (i) a n d (i i) a t (x_{1}, y_{1})$ $\Rightarrow \frac{y_{1}}{n (1 - n - d n)} - \frac{y_{1}}{(n + d n) (1 - n)} = \frac{1}{1 - n - d n} - \frac{1}{1 - n}$ $\Rightarrow y_{1} = n (n + d n)$ $d A = \frac{y_{1} d n}{2} = \frac{n (n + d n) d n}{2} = \frac{n^{2} d n}{2}$ $A = \frac{1}{2} \int_{0}^{1} n^{2} d n = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$
	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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