Объяснение:
[tex]\int\limits^4_1 {(\frac{5}{x}-\frac{1}{2\sqrt{x} }+x^2) } \, dx = \int\limits^4_1 {\frac{5}{x} } \, dx-\int\limits^4_1 {\frac{1}{2\sqrt{x} } } \, dx +\int\limits^4_1 {x^2} \, dx=\\[/tex]
[tex]= 5*\int\limits^4_1 {\frac{dx}{x} } -\frac{1}{2}*\int\limits^4_1 {x^{-\frac{1}{2}} } \, dx + \int\limits^4_1 {x^2} \, dx =5*ln|x|\ |_1^4 -\frac{1}{2} *2*x^{\frac{1}{2}}\ |_1^4 +\frac{x^3}{3}\ |_1^4=\\ =5*(ln4-ln1)-\sqrt{x} \ |_1^4+\frac{4^3}{3} -\frac{1^3}{3} =5*(ln4-0)-(\sqrt{4}-\sqrt{1})+\frac{64}{3}-\frac{1}{3} = \\ =5ln4-(2-1)+\frac{63}{3} =5ln4-1+21=5ln4+20\approx27.[/tex]
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Объяснение:
[tex]\int\limits^4_1 {(\frac{5}{x}-\frac{1}{2\sqrt{x} }+x^2) } \, dx = \int\limits^4_1 {\frac{5}{x} } \, dx-\int\limits^4_1 {\frac{1}{2\sqrt{x} } } \, dx +\int\limits^4_1 {x^2} \, dx=\\[/tex]
[tex]= 5*\int\limits^4_1 {\frac{dx}{x} } -\frac{1}{2}*\int\limits^4_1 {x^{-\frac{1}{2}} } \, dx + \int\limits^4_1 {x^2} \, dx =5*ln|x|\ |_1^4 -\frac{1}{2} *2*x^{\frac{1}{2}}\ |_1^4 +\frac{x^3}{3}\ |_1^4=\\ =5*(ln4-ln1)-\sqrt{x} \ |_1^4+\frac{4^3}{3} -\frac{1^3}{3} =5*(ln4-0)-(\sqrt{4}-\sqrt{1})+\frac{64}{3}-\frac{1}{3} = \\ =5ln4-(2-1)+\frac{63}{3} =5ln4-1+21=5ln4+20\approx27.[/tex]