[tex]f(x)=x^{2}+\frac{x}{\sqrt{x^{2}+9}}[/tex]
a)
[tex]\int\limits^1_0 {x^2} \, dx =\frac{x^3}{3}\ |_0^1=\frac{1}{3}[/tex]
(vezi tabelul de integrale din atasament)
b)
[tex]\int\limits^4_0 {x^2+\frac{x}{\sqrt{x^2+9} } -x^2+\frac{x}{\sqrt{x^2+9} } \, dx =\\\\=\int\limits^4_0 {\frac{2x}{\sqrt{x^2+9} } \, dx[/tex]
[tex]\int\limits^4_0 {\frac{2x}{\sqrt{x^2+9} } \, dx=2\sqrt{x^2+9}\ |_0^4=2\sqrt{16+9}-2\sqrt{0+9} =10-6=4[/tex]
c)
[tex]\int\limits^a_4 {x+\frac{1}{\sqrt{x^2+9} } } \, dx[/tex]
Desfacem in doua integrale si obtinem:
[tex]\frac{x^2}{2}\ |_4^a+ln(x+\sqrt{x^2+9}) \ |_4^a=\frac{a^2}{2}-8+ln(a+\sqrt{a^2+9} ) -ln9=\frac{a^2}{2}-8+ln\frac{a+\sqrt{a^2+9} }{9}[/tex]
[tex]\frac{a^2}{2}-8+ln\frac{a+\sqrt{a^2+9} }{9} =10+ln\frac{a+\sqrt{a^2+9} }{9}\\\\\frac{a^2}{2}-8=10\\\\\frac{a^2}{2}=18\\\\a^2=36\\a=6\\\\a=-6 < 4\ NU[/tex]
a=6
Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/8885003
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