Ответ:
[tex]\frac{1}{6}[/tex] [ед²].
Объяснение:
[tex]S=\int\limits^1_0 {(x+1)} \, dx -\int\limits^1_0 {(x^2+1)} \, dx =\int\limits^1_0 {(x-x^2)} \, dx=(\frac{1}{2} x^2-\frac{1}{3} x^3)|^1_0=\frac{1}{2} -\frac{1}{3} =\frac{1}{6}[unit^2].[/tex]
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Ответ:
[tex]\frac{1}{6}[/tex] [ед²].
Объяснение:
[tex]S=\int\limits^1_0 {(x+1)} \, dx -\int\limits^1_0 {(x^2+1)} \, dx =\int\limits^1_0 {(x-x^2)} \, dx=(\frac{1}{2} x^2-\frac{1}{3} x^3)|^1_0=\frac{1}{2} -\frac{1}{3} =\frac{1}{6}[unit^2].[/tex]