[tex]f(x)=x^{2}+1[/tex]
a)
Vezi tabelul de integrale din atasament
[tex]\int\limits^3_0 {x^2+1} \, dx =\frac{x^3}{3}|_0^3+x|_0^3=\frac{27}{3} +3=9+3=12[/tex]
b)
[tex]\int\limits^e_1 {(x^2+1)e^{x^3+3x}} \, dx \\\\Observam\ ca\ (x^3+3x)'=3x^2+3=3(x^2+1)\\\\Stim\ ca\ (e^u)'=u'\cdot e^u\\\\Atunci\ \int\limits^e_1 {(x^2+1)e^{x^3+3x}} \, dx=\frac{1}{3}e^{x^3+3x}|_0^1=\frac{1}{3}e^4-\frac{1}{3} =\frac{e^4-1}{3}[/tex]
c)
Luam prima integrala si o integram prin parti
[tex]\int\limits^1_0 {(x^2+1)^7} \, dx \\\\f=(x^2+1)^7\ \ \ f'=14x(x^2+1)^6\\\\g'=1\ \ \ \ \ \ \ \ \ \ \ g=x\\\\\int\limits^1_0 {(x^2+1)^7} \, dx=x(x^2+1)^7|_0^1-\int\limits^1_0 {14x^2(x^2+1)^6} \, dx[/tex]
[tex]\int\limits^1_0 {(x^2+1)^7} \, dx=2^7-14\int\limits^1_0 {(x^2+1-1)(x^2+1)^6} \, dx=1128-14\int\limits^1_0 {(x^2+1)^7} \, dx +14\int\limits^1_0 {(x^2+1)^6} \, dx \\\\15\int\limits^1_0 {(x^2+1)^7} \, dx-14\int\limits^1_0 {(x^2+1)^6} \, dx =128[/tex]
Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9928407
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