[tex]f(x)=\frac{x^{2}}{x^{4}+1}[/tex]
a)
Vezi tabelul de integrale din atasament
[tex]\int\limits^1_0 {x^2} \, dx =\frac{x^3}{3} |_0^1=\frac{1}{3}-0= \frac{1}{3}[/tex]
b)
[tex]\int\limits^1_0 {\frac{x^2}{x^4+1} } \, dx[/tex]
[tex]x\in (0,+\infty)\\\\\int\limits^1_0 {\frac{x^2}{x^4+1} } \, dx\leq \int\limits^1_0 {\frac{x}{x^4+1} } \, dx[/tex]
[tex]\int\limits^1_0 {\frac{x}{x^4+1} } \, dx[/tex]
Facem prin schimbare de variabila
[tex]t=x^2\\\\dt=2xdx\\\\x=0\\ t=0\\\\x=1\\ t=1[/tex]
[tex]\int\limits^1_0 {\frac{x}{x^4+1} } \, dx=\frac{1}{2} \int\limits^1_0\frac{1}{t^2+1} \ dt=\frac{1}{2} arctg\ t|_0^1=\frac{1}{2}(arctg1- arctg0)=\frac{\pi}{8}[/tex]
c)
Primitiva lui f este F
F'(x)=f(x)
F(1)=0
[tex]\int\limits^1_0 {F(x)} \, dx[/tex]
Integram prin parti:
[tex]f=F(x)\ \ \ f'=f(x)\\\\g'=1\ \ \ \ g=x[/tex]
[tex]\int\limits^1_0 {F(x)} \, dx=xF(x)|_0^1-\int\limits^1_0xf(x)\ dx=F(1)-\int\limits^1_0\frac{x^3}{x^4+1} \ dx=0-\frac{1}{4}ln(x^4+1)|_0^1=-\frac{1}{4}ln2[/tex]
Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9919077
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