Вычислите интеграл: 6. a) 0 1 10,5*dr; 6) fexdx; B) $$2dx; r) 0 в) [2* г) -2 1 [ 3* dx. 2

Ответ: а) 0.75, б) (e^2-e^(-4))/2, в) 3/ln(2), г) (3^(1/2)-3^(-1/2))/ln(3)

Решение:

а) \[\int_{0}^{1} 0.5x dx = 0.5 \int_{0}^{1} x dx = 0.5 \cdot \frac{x^2}{2} \Biggr|_{0}^{1} = 0.5 \cdot \frac{1^2}{2} - 0.5 \cdot \frac{0^2}{2} = 0.5 \cdot \frac{1}{2} = 0.25\]

б) \[\int_{-2}^{1} e^{2x} dx = \frac{1}{2} e^{2x} \Biggr|_{-2}^{1} = \frac{1}{2} e^{2\cdot 1} - \frac{1}{2} e^{2\cdot (-2)} = \frac{1}{2} e^{2} - \frac{1}{2} e^{-4} = \frac{e^2 - e^{-4}}{2}\]

в) \[\int_{-1/2}^{1} 2^x dx = \frac{2^x}{\ln 2} \Biggr|_{-1/2}^{1} = \frac{2^1}{\ln 2} - \frac{2^{-1/2}}{\ln 2} = \frac{2 - 2^{-1/2}}{\ln 2} = \frac{2 - \frac{1}{\sqrt{2}}}{\ln 2} = \frac{2\sqrt{2} - 1}{\sqrt{2} \ln 2}\]

г) \[\int_{1/2}^{1} 3^x dx = \frac{3^x}{\ln 3} \Biggr|_{1/2}^{1} = \frac{3^1}{\ln 3} - \frac{3^{1/2}}{\ln 3} = \frac{3 - \sqrt{3}}{\ln 3}\]

Ответ: а) 0.25, б) (e^2-e^(-4))/2, в) (2 - 2^(-1/2))/ln(2), г) (3 - sqrt(3))/ln(3)