Задача 57810 Интеграл вычислить,2 примера....

5f88950e258f913735c2ec60

09.02.2021 21:37:28

Интеграл вычислить,2 примера.

5ff1d32e960c3c35933d951f

09.02.2021 23:19:55

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[red]1)[/red][m]\int_1^3{(\frac{3}{x}+x^2)dx}=\int_1^3{\frac{3dx}{x}}+\int_1^3{x^2dx}=3\int_1^3{\frac{dx}{x}}+\int_1^3{x^2dx}=3ln(x)_1^3+\frac{x^3}{3}|_1^3=3(ln(3)-ln(1))+(\frac{3^3}{3}-\frac{1^3}{3})=3(ln(3)-0)+(\frac{27}{3}-\frac{1}{3})=3ln(3)+\frac{26}{3}[/m]
[red]2)[/red][m]sin^2x=\frac{1-cos(2x)}{2}[/m]
[m]\int_0^\frac{\pi}{2}sin^2xdx=\int_0^\frac{\pi}{2}{\frac{1-cos(2x)}{2}dx}=\frac{1}{2}\int_0^\frac{\pi}{2}{(1-cos(2x))dx}=\frac{1}{2}\int_0^\frac{\pi}{2}dx-\frac{1}{2}\int_0^\frac{\pi}{2}cos(2x)dx=\frac{1}{2}\int_0^\frac{\pi}{2}dx-\frac{1}{4}\int_0^\frac{\pi}{2}cos(2x)d(2x)=\frac{x}{2}|_0^\frac{\pi}{2}-\frac{sin(2x)}{4}|_0^\frac{\pi}{2}=\frac{1}{2}(\frac{\pi}{2}-0)-\frac{1}{4}(sin(2*\frac{\pi}{2})-sin(2*0))=\frac{1}{2}*\frac{\pi}{2}-\frac{1}{4}*0=\frac{\pi}{4}[/m]