Aplicatii ale integralei definite Aria unei suprafete plane

Dupa ce am invatat sa calculam integralele definite cu ajutorul mai multor metode,a venit vremea sa stim unde putem sa le aplicam in practica. Astfel,astazi discutam despre aplicatii ale integralei definite, aria unei suprafete plane si
aria unei suprafete marginita de grafice de functii .
Incepem prin a enunta o teorema cu ajutorul careia o sa rezolvam mai multe exercitii:
Fie functiile $f,g:\left[a,b\right]\rightarrow R$ continue sau continue pe portiuni. Presupunem ca $g\left(x\right)\leq f\left(x\right), \forall x\in \left[a, b\right]$ . Suprafata dintre graficele functiilor f si g pe $\left[a,b\right]$ este
$\Gamma=\left\{\left(x,y\right)|a\leq x\leq b\;\; sig\left(x\right)\leq y\leq f\left(x\right)\right\}$ .
Aria suprafetei este $\Gamma$ este
$\int^{a}_{b}\left(f\left(x\right)-g\left(x\right)\right)dx$
In particular daca $f:\left[a,b\right]\rightarrow R$ este continua si $f\left(x\right) \geq 0, \forall x\in\left[a,b\right]$ , atunci suprafata
$\Gamma{f}=\left\{\left(x, y\right)\in R^{2}|a\leq x\leq b\;\; si \;\; 0\leq y\leq f\left(x\right)\right\}$ are arie si $A\left(\Gamma_{f}\right)=\int^{b}_{a}f\left(x\right)dx$
Exercitiu
1) Sa se determine aria suprafetei cuprinse intre graficul functiei f si axa Ox pentru
a) $f:\left[0,1\right]\rightarrow R, f\left(x\right)=\frac{1}{x^{2}+4x+8}$
Solutie
$f\left(x\right)=\frac{1}{x^{2}+4x+8}\geq 0, \forall x\in\left[0,1\right]\Rightarrow$
$A\left(\Gamma_{f}\right)=\int^{1}_{0}\frac{1}{x^{2}+4x+8} dx=$
Ca sa rezolvam integrala de mai sus calculam
$x^{2}+4x+8=0 \\ \Delta=b^{2}-4\cdot a\cdot c=16-4\cdot 1\cdot 8=16-32=-16<0$
Deci pentru a rezolva integrama de mai sus scriem:

$\int^{1}_{0}\frac{1}{x^{2}+4x+8} dx=\int^{1}_{0}\frac{1}{x^{2}+4x+4+4}dx=\int^{1}_{0}\frac{1}{\left(x+2\right)^{2}+2^{2}}=\frac{1}{2}arctan\frac{x+2}{2}|^{1}_{0}=\frac{1}{2}\left(arctan\frac{1+2}{2}-arctan\frac{0+2}{2}\right)=\frac{1}{2}\left(arctan\frac{3}{2}-arctan\frac{2}{2}\right)=\frac{1}{2}\left(arctan\frac{3}{2}-\frac{\pi}{4}\right)=\frac{1}{2}arctan\frac{3}{2}-\frac{\pi}{8}$ .
b) $f:\left[1, e\right]\rightarrow R, f\left(x\right)=\ln x$
Functia
$f\left(x\right)=\ln x\geq 0,\;\; \forall x\in \left[1, e\right]\Rightarrow$
$A\left(\Gamma_{f}\right)=\int^{e}_{1}\ln x dx=\int^{e}_{1} x^{'}\cdot \ln x dx=$
$x\cdot\ln x|^{e}_{1}-\int^{e}_{1}x\left(\ln x\right)^{'}dx=$
$e\cdot \ln e-1\ln 1-\int^{e}_{1}x\cdot\frac{1}{x}dx=$
$e-1\cdot 0-\int^{1}_{e} dx=e-x|^{e}_{1}=1$
c) $f,g:\left[0,4\right]\rightarrow R, f\left(x\right)=-\sqrt{x}, g\left(x\right)=\sqrt{x}$
Solutie
Fie $h:\left[0,4\right]\rightarrow R, h\left(x\right)=f\left(x\right)-g\left(x\right)=-\sqrt{x}-\sqrt{x}=-2\sqrt{x}<0\;\; \forall x\in\left[0,4\right]\Rightarrow f\left(x\right)\leq g\left(x\right), \forall x\in \left[0, 4\right]$
Atunci avem
$A\left(\Gamma{f,g}\right)=\int^{4}_{0}\left(g\left(x\right)-f\left(x\right)\right)dx=$
$\int^{4}_{0}\left(\sqrt{x}-\left(-\sqrt{x}\right)\right)dx=\int^{4}_{0}\left(\sqrt{x}+\sqrt{x}\right)dx=$
$\int^{4}_{0} 2\cdot \sqrt{x} dx=2\int^{4}_{0}\sqrt{x} dx=2\int^{4}_{0}x^{\frac{1}{2}}dx=$
$2\cdot \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}|^{4}_{0}=$
$2\cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}}|^{4}_{0}=$
$2\cdot \frac{2}{3}\cdot x^{\frac{3}{2}}|^{4}_{0}=$
$2\cdot \frac{2}{3}\cdot \sqrt{x^{3}}|^{4}_{0}=\frac{2}{3}\left(\sqrt{4^{3}}-\sqrt{0^{3}}\right)=$
$2\cdot \frac{2}{3}\cdot 4\sqrt{4}=\frac{2}{3}\cdot 4\cdot 2=$
$2\cdot \frac{2}{3}\cdot 8=2\cdot \frac{16}{3}=\frac{32}{3}$

2) Calculati aria multimii plane:
$A=\left\{\left(x,y\right)|-1\leq x\leq 1\;\; si\;\; \frac{x^{2}}{2}\leq y\leq \frac{1}{1+x^{2}}\right\}$
Solutie
Conform primei teoreme pe care am enuntat-o avem ca
1, si -1 sunt capetele intervalului, adica avem functiile f si g
$f, g:\left[-1, 1\right]\rightarrow R, g\left(x\right)=\frac{1}{1+x^{2}}\;\; si f\left(x\right)=\frac{x^{2}}{2}$ , astfel avem din ipoteza ca functia
$f\left(x\right)<g\left(x\right),\;\;\forall x\in \left[-1, 1\right]$
Astfel
$A\left({\Gamma_{f,g}}\right)=\int^{1}_{-1}\left(g\left(x\right)-f\left(x\right)\right)dx=\int^{1}_{-1}\left(\frac{1}{1+x^{2}}-\frac{x^{2}}{2}\right)dx=\int^{1}_{-1}\left(\frac{2-\left(1+x^{2}\right)\cdot x^{2}}{2\left(1+x^{2}\right)}\right) dx= \int^{1}_{-1}\frac{2-x^{2}-x^{4}}{2\left(1+x^{2}\right)}dx=\int^{1}_{-1}\frac{2}{2\left(1+x^{2}\right)}dx-\int^{1}_{-1}\frac{x^{2}\left(1+x^{2}\right)}{2\left(1+x^{2}\right)}dx=\int^{1}_{-1}\frac{1}{1+x^{2}}dx-\int^{1}_{-1}x^{2}dx=\frac{1}{1}arctan{x}{1}|^{1}_{-1}-\frac{x^{2+1}}{2+1}|^{1}_{-1}=arctan{1}-arctan{-1}-\frac{x^{3}}{3}|^{1}_{-1}=\frac{\pi}{4}-\frac{-\pi}{4}-\frac{1}{3}+\frac{\left(-1\right)^{3}}{3}=\frac{\pi}{4}+\frac{\pi}{4}-\frac{1}{3}-\frac{1}{3}=\frac{2\pi}{4}-\frac{2}{3}=\frac{\pi}{2}-\frac{2}{3}$

Aplicatii ale integralei definite Aria unei suprafete plane

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