Question

вычислите интегралы
35 БАЛЛОВ

Answer 1

Решение:

$1) \int\limits^2_0 {(x^{2}+x-2)} \, dx = \int\limits {(x^{2}+x-2)} \, dx = \int\limits {x^{2}} \, dx +\int\limits {x} \, dx -\int\limits {2} \, dx = (\frac{x^{3}}{3}+\frac{x^{2}}{2}-2x)|^2_0 = \frac{2^{3}}{3}+\frac{2^{2}}{2}-2*2-(\frac{0^{3}}{3}+\frac{0^{2}}{2}-2*0) = \frac{8}{3}+\frac{4}{2}-4-(\frac{0}{3}+\frac{0}{2}-0) = \frac{8}{3}+2-4-0-0+0 = \frac{8}{3}-2 = \frac{8-6}{3} = \frac{2}{3}$

$2) \int\limits^0_{-2} {(3-2x-x^{2})} \, dx = \int\limits {(3-2x-x^{2})} \, dx = \int\limits {3} \, dx - \int\limits {2x} \, dx - \int\limits {x^{2}} \, dx = 3x-2*\frac{x^{2}}{2}-\frac{x^{3}}{3} = (3x-x^{2}-\frac{x^{3}}{3})|^0_{-2} = 3*0-0^{2}-\frac{0^{3}}{3}-(3*(-2)-(-2)^{2}-\frac{(-2)^{3}}{3}) = 0-0-0-(-6-4-\frac{-8}{3}) = 0+6+4-\frac{8}{3}=10-\frac{8}{3} = \frac{30-8}{3} = \frac{22}{3} = 7\frac{1}{3}$

$3) \int\limits^3_0 {(3x^{2}+x-2)} \, dx = \int\limits {(3x^{2}+x-2)} \, dx = \int\limits {3x^{2}} \, dx +\int\limits {x} \, dx - \int\limits {2} \, dx = 3*\frac{x^{3}}{3}+\frac{x^{2}}{2}-2x = (x^{3}+\frac{x^{2}}{2}-2x)|^3_0 = 3^{3}+\frac{3^{2}}{2}-2*3-(0^{3}+\frac{0^{2}}{2}-2*0) = 27+\frac{9}{2}-6-(0+0-0) = 27+4,5-6-0 = 25,5$

$4) \int\limits^3_1 {(3-x^{2})} \, dx = \int\limits {(3-x^{2})} \, dx = \int\limits {3} \, dx - \int\limits {x^{2}} \, dx = 3\int\limits\, dx -\frac{x^{2+1}}{2+1} = (3x-\frac{x^{3}}{3})|^3_1 = 3*3-\frac{3^{3}}{3}-(3*1-\frac{1^{3}}{3}) = 9-\frac{27}{3}-(3-\frac{1}{3}) = 9-9-3+\frac{1}{3} = \frac{-9+1}{3} = \frac{-8}{3} = -\frac{8}{3} = -2\frac{2}{3}$

$5) \int\limits^{\frac{\pi}{2}}_0 {9cosx} \, dx = \int\limits {9cosx} \, dx = 9\int\limits{cosx} \, dx = 9*sinx = 9sinx|^{\frac{\pi}{2}}_0 = 9sin\frac{\pi}{2} - 9sin0 = 9*1-9*0 = 9-0 = 9$

$6) \int\limits^{\frac{\pi}{4}} _0 {\frac{2dx}{cos^{2}x}} = \int\limits {\frac{2dx}{cos^{2}x}} = 2\int\limits {\frac{1dx}{cos^{2}x}} = 2*tgx = 2tgx|^{\frac{\pi}{4}}_0 = 2tg\frac{\pi}{4}-2tg0 = 2*1-2*0 = 2-0 = 2$

Ответ: $1)\frac{2}{3} ; \\2) 7\frac{1}{3} ; \\3)25,5 ; \\4) -2\frac{2}{3} ; \\5) 9 ; \\6)2$

Решено от : $DK954$