Question

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

5p a) Arătați că $\int_{0}^{\pi} \frac{f(x)}{e^{x}} d x=0$

\begin{tabular}{l|l}
\hline $5 p$ & b) Calculați $\int_{0}^{\frac{\pi}{2}} f(x) d x .$ \\
$5 p$& c) Arătați că $\int_{0}^{\frac{\pi}{3}} \frac{f\left(x+\frac{\pi}{2}\right)}{f(x)} d x=-e^{\frac{\pi}{2}} \ln 2 .$
\end{tabular}

Answer 1

$f(x)=e^{x} \cos x$

a)

Vezi tabelul de integrale din atasament

$\int\limits^{\pi}_0 {cosx} \, dx =sinx|_0^{\pi}=sin\pi-sin0=0$

b)

Integram prin parti si notam integrala noastra cu I

$f=cosx\ \ \ f'=-sinx\\\\g'=e^x\ \ \ g=e^x\\\\I=e^xcosx|_0^{\frac{\pi}{2} }+\int\limits^{\frac{\pi}{2}} _0 {e^xsinx} \, dx$

Calculam separat integrala $\int\limits^{\frac{\pi}{2}} _0 {e^xsinx} \, dx$

$f=sinx\ \ \ f'=cosx\\\\g'=e^x\ \ \ g=e^x\\\\\int\limits^{\frac{\pi}{2}} _0 {e^xsinx} \, dx =e^xsinx|_0^{\frac{\pi}{2}}- \int\limits^{\frac{\pi}{2}} _0e^xcosx=e^xsinx|_0^{\frac{\pi}{2}}-I$

$I=e^xcosx|_0^{\frac{\pi}{2}}+e^xsinx|_0^{\frac{\pi}{2}}-I\\\\2I=e^xcosx|_0^{\frac{\pi}{2}}+e^xsinx|_0^{\frac{\pi}{2}}\\2I=e^{\frac{\pi}{2}}-1\\\\I=\frac{e^{\frac{\pi}{2}}-1}{2}$

c)

$\int\limits^{\frac{\pi}{3} }_0{\frac{e^{x+\frac{\pi}{2}}cos(x+\frac{\pi}{2}) }{e^xcosx} } \, dx =\int\limits^{\frac{\pi}{3} }_0\frac{e^x\cdot e^{\frac{\pi}{2}}(-sinx) }{e^xcosx} \ dx=-e^{\frac{\pi}{2}}\cdot \int\limits^{\frac{\pi}{3} }_0 \frac{sinx}{cosx} =e^{\frac{\pi}{2}}\cdot ln(cosx)|_0^{\frac{\pi}{3}} =e^{\frac{\pi}{2}}(ln\frac{1}{2}-ln1)=-e^{\frac{\pi}{2}}ln2$

Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9928393

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