[tex]f(x)=x+2 \ln (2 x+1)[/tex]
a)
[tex]\int\limits^1_0 {x} \, dx=\frac{x^2}{2}|_0^1=\frac{1}{2}[/tex]
Vezi tabelul de integrale din atasament
b)
[tex]\int\limits^1_0 {x+2ln(2x+1)} \, dx =\frac{1}{2}+ 2\int\limits^1_0 {ln(2x+1)} \, dx=\frac{1}{2}+2(ln3-1+\frac{ln3}{2} )=\frac{1}{2}+2ln3-2+ln3=3ln3-\frac{3}{2}[/tex]
Luam integrala si o integram prin parti
[tex]\int\limits^1_0 {ln(2x+1)} \, dx[/tex]
[tex]f=ln(2x+1)\ \ \ f'=\frac{2}{2x+1} \\\\g'=1\ \ \ \ \ \ \ \ g=x\\\\\int\limits^1_0 {ln(2x+1)} \, dx=xln(2x+1)|_0^1-\int\limits^1_0\frac{2x}{2x+1} \ dx=ln3-\int\limits^1_01\ dx+\int\limits^1_0\frac{1}{2x+1}\ dx=ln3-1+\frac{1}{2}ln(2x+1)|_0^1=\\\\ =ln3-1+\frac{ln3}{2}[/tex]
c)
F este primitiva lui f
F'(x)=f(x)
Facem monotonia functiei F
F'(x)≥0⇒ F este crescatoare pe [0,+∞)
[tex]\pi\leq \frac{16}{5} \\\\F(\pi)\leq F(\frac{16}{5})[/tex]
Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9918932
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