Question Number 4535 by FilupSmith last updated on 05/Feb/16
$$\mathrm{Lets}\:\mathrm{say}\:\mathrm{we}\:\mathrm{have}\:\mathrm{three}\:\mathrm{points}: \\ $$$${A}\left(\mathrm{0},\:\mathrm{0}\right) \\ $$$${B}\left({x},\:{y}\right) \\ $$$${C}\left(\delta{x},\:\delta{y}\right) \\ $$$$ \\ $$$$\mathrm{Assuming}\:\mathrm{that}\:\mathrm{both}\:{B}\:\mathrm{and}\:{C}\:\mathrm{are}\:\mathrm{point} \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{fuction}\:{y}={f}\left({x}\right),\:\mathrm{we}\:\mathrm{can}\:\mathrm{calculate} \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{under}\:\mathrm{the}\:\mathrm{point}\:\mathrm{where}\:\mathrm{it}\:\mathrm{makes} \\ $$$$\mathrm{a}\:\mathrm{right}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{and}\:\mathrm{x}−\mathrm{axis}. \\ $$$$ \\ $$$$\mathrm{Can}\:\mathrm{we}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:{ABC}? \\ $$
Commented by FilupSmith last updated on 05/Feb/16
Commented by Yozzii last updated on 07/Feb/16
![tanα=y/x, tanφ=δy/δx ⇒α−φ=tan^(−1) (y/x)−tan^(−1) (δy/δx) A_(△ABC) =(1/2)(√((y^2 +x^2 )({δy}^2 +{δx}^2 )))[sin{tan^(−1) (y/x)−tan^(−1) ((δy)/(δx))}]](Q4566.png)
$${tan}\alpha={y}/{x},\:{tan}\phi=\delta{y}/\delta{x} \\ $$$$\Rightarrow\alpha−\phi={tan}^{−\mathrm{1}} \left({y}/{x}\right)−{tan}^{−\mathrm{1}} \left(\delta{y}/\delta{x}\right) \\ $$$${A}_{\bigtriangleup{ABC}} =\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\left({y}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\left(\left\{\delta{y}\right\}^{\mathrm{2}} +\left\{\delta{x}\right\}^{\mathrm{2}} \right)}\left[{sin}\left\{{tan}^{−\mathrm{1}} \frac{{y}}{{x}}−{tan}^{−\mathrm{1}} \frac{\delta{y}}{\delta{x}}\right\}\right] \\ $$