[tex]f(x)=\frac{2 x+1}{x+1}[/tex]
a)
[tex]\int_{0}^{1}(x+1) f(x) d x=\int_{0}^{1}(x+1)\times \frac{2x+1}{x+1} d x=\int_{0}^{1}(2x+1) d x\\\\\int_{0}^{1}(2x+1) d x=\int_{0}^{1}2x\ d x+\int_{0}^{1}1\ d x=\frac{2x^2}{2}\ |_0^1+x\ |_0^1=1+1=2[/tex]
Nota: am desfacut integrala in doua integrale si am aplicat formula din tabelul de integrale
b)
[tex]\int\limits^1_0 {\frac{2x+1}{x+1} } \, dx[/tex]
Facem un artificiu de calcul, il scriem pe 1=2-1 si apoi dam factor comun pe 2 , desfacem fractia in doua fractii si obtinem:
[tex]\int\limits^1_0 {\frac{2x+2-1}{x+1} } \, dx=\int\limits^1_0 {\frac{2(x+1)-1}{x+1} } \, dx=\int\limits^1_0 {\frac{2(x+1)}{x+1} } \, dx-\int\limits^1_0 {\frac{1}{x+1} } \, dx[/tex]
Il scriem pe 1=x'
[tex]\int\limits^1_0 {\frac{2(x+1)}{x+1} } \, dx-\int\limits^1_0 {\frac{x'}{x+1} } \, dx=\int\limits^1_0 {2} \, dx -ln|x+1|\ |_0^1=2x\ |_0^1-ln2=2-ln2[/tex]
c)
[tex]I_{n}=\int_{0}^{1} e^{x}(x+1)^{n}(f(x))^{n} d x[/tex]
[tex]I_{n}=\int_{0}^{1} e^{x}(x+1)^{n}(\frac{2x+1}{x+1} )^{n}\ d x\\\\I_{n}=\int_{0}^{1} e^{x}(2x+1)^n\ d x[/tex]
Integram prin parti:
[tex]f(x)=(2x+1)^n\ \ \ \ \ \ \ \ \ f'(x)=2n(2x+1)^{n-1}\\\\\\g'(x)=e^x\ \ \ \ \ \ \ \ \ \ \ \ g(x)=e^x[/tex]
[tex]I_n=e^x(2x+1)^n\ |_0^1-2nI_{n-1}\\\\I_n=e\times 3^n-1-2nI_{n-1}\\[/tex]
De aici rezulta ca:
[tex]I_n+2nI_{n-1}=3^n\times e-1[/tex]
Un exercitiu similar de bac gasesti aici: https://brainly.ro/tema/4506298
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