Question

Se consideră funcţia $f:(0,+\infty) \rightarrow \mathbb{R}, f(x)=\ln x$.

$5 \mathbf{p}$ a) Arătaţi că $\int_{1}^{e} f^{\prime}(x) d x=1$.

$5 \mathbf{p}$ b) Calculați $\int_{1}^{e} \frac{f^{2}(x)}{x} d x$.

$5 p$ c) Determinaţi numărul real $p, p\ \textgreater \ 1$, ştiind că $\int_{1}^{p} x f(x) d x=\frac{p^{2}}{2} \ln p-\frac{3}{4}$.

Answer 1

$f(x)=\ln x$

a)

$\int\limits^e_1 {f'(x)} \, dx =f(x)|_1^e\\\\\int\limits^e_1 {\frac{1}{x}}\ dx=lnx|_1^e=lne-ln1=1-0=1$

b)

$\int\limits^e_1 {\frac{ln^2x}{x} } \, dx =\int\limits^e_1 {ln^2x\cdot\frac{1}{x} } \, dx$

Il scriem pe $\frac{1}{x}=(lnx)'$

$\int\limits^e_1 {ln^2x\cdot(lnx)'} \, dx =\frac{1}{3} ln^3x|_1^e=\frac{1}{3}$

c)

Mai intai calculam urmatoarea integrala:

$\int\limits^p_1 {xlnx} \, dx$

O calculam prin integrarea prin parti

$f=lnx\ \ \ \ \ \ \ f'=\frac{1}{x} \\\\g'=x\ \ \ \ \ \ \ \ g=\frac{x^2}{2} \\\\\int\limits^p_1 {xlnx} \, dx =\frac{x^2}{2}lnx|_1^p-\int\limits^p_1 {\frac{x^2}{2}\cdot\frac{1}{x} } \, dx =\frac{x^2}{2}lnx|_1^p-\int\limits^p_1 {\frac{x}{2} } \, dx =\\\\=\frac{x^2}{2}lnx|_1^p-\int\limits^p_1 {\frac{x}{2} } \, dx =\frac{x^2}{2}lnx|_1^p-\frac{x^2}{4}|_1^p=\frac{p^2}{2}lnp-0-\frac{p^2}{4}+\frac{1}{4}$

Apoi egalam:

$\frac{p^2}{2}lnp-0-\frac{p^2}{4}+\frac{1}{4} =\frac{p^2}{2}lnp-\frac{3}{4} \\\\\\\\\frac{p^2}{4}=\frac{4}{4} =1\\\\p^2=4\\\\p=-2 < 1\ NU\\\\p=2$

p=2 este solutia finala

Un exercitiu similar de bac il gasesti aici: https://brainly.ro/tema/5727603

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