[tex]f(x)=x \sqrt{x^{2}+1}[/tex]
a)
[tex]\int\limits^e_1{\frac{\sqrt{x^2+1} }{x\sqrt{x^2+1} } } \, dx =\int\limits^e_1 {\frac{1}{x} } \, dx =lnx|_1^e=lne-ln1=1-0=1[/tex]
(vezi tabel integrale in atasament)
b)
[tex]\int\limits^2_1 {x^2(x^2+1)} \, dx =\int\limits^2_1 x^4+x^2\ dx=\int\limits^2_1x^4\ dx+\int\limits^2_1x^2\ dx=\frac{x^5}{5}|_1^2+\frac{x^3}{3}|_1^2=\\=\frac{32}{5}-\frac{1}{5}+\frac{8}{3}-\frac{1}{3} =\frac{31}{5}+\frac{7}{3}=\frac{128}{15}[/tex]
c)
a≥2020
Pentru x≥0, f(x)≥0
Deci [tex]\int\limits^a_{2020} {f(x)} \, dx \geq 0[/tex]
[tex]\int\limits^a_0 {f(x)} \, dx -\int\limits^{2020}_0 {f(x)} \, dx =\int\limits^{2020}_0 {f(x)} \, dx+\int\limits^a_{2020} {f(x)} \, dx -\int\limits^{2020}_0 {f(x)} \, dx=\int\limits^a_{2020} {f(x)} \ dx[/tex]
Am demonstrat mai sus ca integrala aceea este ≥0
Deci [tex]\int\limits^a_{2020} {f(x)} \geq 0\\\\\int\limits^{2020}_0 {f(x)} \leq \int\limits^a_0{f(x)}[/tex]
Un alt exercitiu similar de bac il gasesti aici: https://brainly.ro/tema/4786722
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