Задача 75883 Вычислите определение интеграл...
Условие
Решение
1) [m]\int_{-1}^1 (1 - 5x^4) dx = (x - \frac{5x^5}{5}) |_{-1}^1 = (x - x^5)|_{-1}^1 =[/m]
[m]= (1 - 1^5) - (-1 - (-1)^5) = 0 - 0 = 0[/m]
2) [m]\int_{1}^4 (\frac{4}{x^2} + 2x - 3x^2) dx = (-\frac{4}{x}+x^2-x^3)|_{1}^4 =[/m]
[m]=(-\frac{4}{4}+4^2-4^3) - (-\frac{4}{1}+1^2-1^3) =[/m]
[m]= -1+16-64+4-1+1 = -45[/m]
3) [m]\int_{0}^1 (2x - 1)^2 dx = \frac{1}{2} \cdot \frac{(2x-1)^3}{3}|_{0}^1 =\frac{1}{2} \cdot (\frac{(2 \cdot 1-1)^3}{3} - \frac{(2 \cdot 0-1)^3}{3}) =[/m]
[m]= \frac{1}{2} \cdot (\frac{1^3}{3} - \frac{(-1)^3}{3}) =\frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}[/m]
4) [m]\int_{1}^4 (\frac{2}{\sqrt{x}} - \frac{1}{x^2}) dx = (2 \cdot (-\frac{1}{2}) \cdot x^{-3/2} + \frac{1}{x})|_{1}^4 =[/m]
[m]= (-\frac{1}{\sqrt{x^3}} + \frac{1}{x})|_{1}^4 = (-\frac{1}{\sqrt{4^3}} + \frac{1}{4}) - (-\frac{1}{\sqrt{1^3}} + \frac{1}{1}) = -\frac{1}{8} + \frac{1}{4} + 1 - 1 = \frac{1}{8}[/m]