Question

Se consideră funcţia $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=\frac{1-e^{x}}{1+e^{x}}$.

5p a) Determinaţi primitiva $G$ a functiei $g: \mathbb{R} \rightarrow \mathbb{R}, g(x)=\left(1+e^{x}\right) f(x)$ pentru care $G(0)=0$.
$5 p \mid$ b) Calculați $\int_{0}^{1} f(x) d x$
5p c) Demonstrați că $\int_{-1}^{1} f(x) \cos x d x=0$

Answer 1

$f(x)=\frac{1-e^{x}}{1+e^{x}}$

a)

$g(x)=(1+e^x)\cdot \frac{1-e^{x}}{1+e^{x}}=1-e^x$

Primitiva G a functiei g

G'(x)=g(x)

$G(x)=\int\limits {g(x)} \, dx +C$

$G(x)=\int\limits {1-e^x} \, dx \ +C=x-e^x+C\\\\G(0)=0\\\\0-e^0+C=0\\\\-1+C=0\\\\C=1\\\\G(x)=x-e^x+1$

b)

$\int\limits^1_0 { \frac{1-e^{x}}{1+e^{x}}} \, dx =\int\limits^1_0 { \frac{1+e^{x}-2e^{x}}{1+e^{x}}} \, dx =\\\\=\int\limits^1_0 {1- \frac{2e^{x}}{1+e^{x}}} \, dx =[x-2ln(1+e^x)]|_0^1=1-2ln(1+e)-0+2ln2=\\\\=1+2ln\frac{2}{1+e}$

c)

Notam:

$\int\limits^1_{-1} {f(x)cox} \, dx =I$

$I=-\int\limits^{-1}_{1} {f(-x)co(-x)} \, dx\\\\I=-\int\limits^{-1}_{1}\frac{1-e^{-x}}{1+e^{-x}} cosx\ dx\\\\I=-\int\limits^{-1}_{1}\frac{1-\frac{1}{e^x} }{1+\frac{1}{e^x} } cosx\ dx\\\\I=-\int\limits^{-1}_{1}\frac{\frac{e^x-1}{e^x} }{\frac{e^x+1}{e^x} } cosx\ dx$

$I=\int\limits^{1}_{-1}\frac{e^x-1}{e^x+1} cos x\ dx\\\\I=-\int\limits^{1}_{-1}\frac{1-e^x}{e^x+1}cosx\ dx\\\\ I=-I\\\\2I=0\\\\I=0$

Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/1452381

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