[tex]f(x)=3 x^{3}+4 x^{2}[/tex]
a)
Vezi tabelul de integrale din atasament
[tex]\int\limits^2_0 {3x^3} \, dx =\frac{3x^4}{4}|_0^2=\frac{3\cdot 16}{4} -0=12[/tex]
b)
F'(x)=f(x)
F(0)=2020
[tex]F(x)=\int\limits {f(x)} \, dx =\frac{3x^4}{4}+\frac{4x^3}{3}+C\\\\ F(x)=\frac{3x^4}{4}+\frac{4x^3}{3} +C\\\\F(0)=0+0+C=2020\\\\\ C=2020\\\ F(x)=\frac{3x^4}{4}+\frac{4x^3}{3} +2020[/tex]
c)
[tex]\int\limits^m_1 {\frac{3x^3+4x^2}{x^2} } \, dx= \int\limits^m_13x\ dx+\int\limits^m_14 \ dx=\frac{3x^2}{2}|_1^m+4x|_1^m=\frac{3m^2}{2}-\frac{3}{2}+4m-4 \\\\ \frac{3m^2}{2}-\frac{3}{2}+4m-4 =\frac{17}{2}\\\\ 3m^2-3+8m-8=17\\\\ 3m^2+8m-28=0\\\\\Delta=64+336=400\\\\m_1=\frac{-8+20}{6} =2\\\\m_2=\frac{-8-20}{6} < 1\ NU[/tex]
Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9928472
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