Вычислите определенные интегралы: \(\int\)_0^3 x^3 dx, \(\int\)_0^2 (x^2-2x+1) dx, \(\int\)_{-1}^0 (3x^3+4x) dx, \(\int\)_0^{\(\pi\)/2} cos x dx

Решение:

1. \[ \int_0^3 x^3 dx = \left[ \frac{x^4}{4} \right]_0^3 = \frac{3^4}{4} - \frac{0^4}{4} = \frac{81}{4} \]
2. \[ \int_0^2 (x^2-2x+1) dx = \left[ \frac{x^3}{3} - x^2 + x \right]_0^2 = \left( \frac{2^3}{3} - 2^2 + 2 \right) - \left( \frac{0^3}{3} - 0^2 + 0 \right) = \frac{8}{3} - 4 + 2 = \frac{8}{3} - 2 = \frac{8-6}{3} = \frac{2}{3} \]
3. \[ \int_{-1}^0 (3x^3+4x) dx = \left[ \frac{3x^4}{4} + 2x^2 \right]_{-1}^0 = \left( \frac{3(0)^4}{4} + 2(0)^2 \right) - \left( \frac{3(-1)^4}{4} + 2(-1)^2 \right) = 0 - \left( \frac{3}{4} + 2 \right) = -\left( \frac{3+8}{4} \right) = -\frac{11}{4} \]
4. \[ \int_0^{\pi/2} \cos x dx = \left[ \sin x \right]_0^{\pi/2} = \sin(\frac{\pi}{2}) - \sin(0) = 1 - 0 = 1 \]

Ответ:

• \[ \frac{81}{4} \]
• \[ \frac{2}{3} \]
• \[ -\frac{11}{4} \]
• \[ 1 \]