Question

Se consideră funcţia $f:(0,+\infty) \rightarrow \mathbb{R}, f(x)=x^{2}+\ln x$.

$5 p$ a) Arătați că $\int_{1}^{3}(f(x)-\ln x) d x=\frac{26}{3}$.

$5 \mathbf{p}$ b) Calculați $\int_{1}^{2}\left(f(x)-x^{2}\right) d x$.

$5 p$ c) Arătați că $\int_{1}^{2} \frac{1}{x} f\left(\frac{1}{x}\right) d x=\frac{3-4 \ln ^{2} 2}{8}$.

Answer 1

$f(x)=x^{2}+\ln x$

a)

Vezi tabelul de integrale din atasament

$\int\limits^3_1 {x^2} \, dx =\frac{x^3}{3}|_1^3=\frac{27}{3}-\frac{1}{3} =\frac{26}{3}$

b)

$\int\limits^2_1 {lnx} \, dx =$

Integram prin parti

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

$\int\limits^2_1 {lnx} \, dx =xlnx|_1^2-\int\limits^2_1 {1 \, dx =2ln2-ln1-x|_1^2=2ln2-2+1=2ln2-1$

c)

$f(\frac{1}{x})=\frac{1}{x^2}+ln\frac{1}{x} = \frac{1}{x^2}+ln1-lnx=\frac{1}{x^2}-lnx$

$\int\limits^2_1 {\frac{1}{x^3}-\frac{lnx}{x} } \, dx =\int\limits^2_1 {\frac{1}{x^3}} \, dx -\int\limits^2_1 {\frac{lnx}{x} } \, dx =\frac{x^{-2}}{-2} |_1^2-\int\limits^2_1 {lnx\cdot ln'x } \, dx=-\frac{1}{8}+\frac{1}{2}-\frac{1}{2}ln^2x|_1^2=-\frac{1}{8}+\frac{1}{2}-\frac{1}{2}ln^22=\frac{3-4ln^22}{8}$

Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9919079

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