Question

Se consideră funcţia $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=4 x-\frac{2 x}{x^{2}+1}+\frac{1}{x^{2}+1}$.

$5 p$ a) Arătaţi că $\int_{0}^{1}\left(x^{2}+1\right) f(x) d x=3$
[$5 p$ b) Calculați $\int_{0}^{1} f(x) d x$
$5 p$ c) Determinați numărul real a pentru care $\int_{1}^{e}\left(f(x)+\frac{2 x-1}{x^{2}+1}\right) \ln x d x=e^{2}+a$.

Answer 1

$f(x)=4 x-\frac{2 x}{x^{2}+1}+\frac{1}{x^{2}+1}$

a)

$\int\limits^1_0 {(x^2+1)f(x)} \, dx =\int\limits^1_04x(x^2+1)-2x+1\ dx=\int\limits^1_04x^3+2x+1\ dx=\frac{4x^4}{4} |_0^1+x^2|_0^1+x|_0^1=1+1+1=3$

b)

$\int\limits^1_0 {4 x-\frac{2 x}{x^{2}+1}+\frac{1}{x^{2}+1}} \, dx =2x^2|_0^1-ln(x^2+1)|_0^1+arctgx|_0^1=2-ln2+\frac{\pi}{4}$

(Vezi in atasament formulele de integrare si derivare)

c)

$\int\limits^e_1 {(4x-\frac{2x}{x^2+1} +\frac{1}{x^2+1}+\frac{2x-1}{x^2+1})lnx } \, dx=\int\limits^e_1 4x lnx\ dx=4\int\limits^e_1 x lnx\ dx$

Notam $I=\int\limits^e_1 x lnx\ dx$

O integram prin parti

$f=lnx\ \ \ \ f'=\frac{1}{x} \\\\g'=x\ \ \ \ g=\frac{x^2}{2}$

$I=\frac{x^2}{2}lnx|_1^e -\frac{1}{2}\int\limits^e_1 {x} \, dx \\\\I=\frac{x^2}{2}lnx|_1^e -\frac{1}{2}\cdot \frac{x^2}{2}|_1^e \\\\I=\frac{e^2}{2}-\frac{e^2}{4}+\frac{1}{4}$

$4I=4(\frac{e^2}{2}-\frac{e^2}{4}+\frac{1}{4} )\\\\2e^2-e^2+1=e^2+a\\\\a=1$

Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9882193

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