Af 1) ∫₂⁴ xdx 2) ∫₅¹⁰ 4dx Определенный интеграсе 3) ∫₂³ (3x-4)dx 4) ∫₁² 8ˣdx 5) ∫₁(6x²-8x+9)dx 6) ∫₀² (2x-1)³dx 7) ∫₁³ 3⁴⁻²ˣ dx

1) \[\int_{2}^{4} x dx = \frac{x^2}{2} \Big|_{2}^{4} = \frac{4^2}{2} - \frac{2^2}{2} = \frac{16}{2} - \frac{4}{2} = 8 - 2 = 6\]

2) \[\int_{5}^{10} 4 dx = 4x \Big|_{5}^{10} = 4(10) - 4(5) = 40 - 20 = 20\]

3) \[\int_{2}^{3} (3x - 4) dx = \left(\frac{3x^2}{2} - 4x\right) \Big|_{2}^{3} = \left(\frac{3(3^2)}{2} - 4(3)\right) - \left(\frac{3(2^2)}{2} - 4(2)\right) = \left(\frac{27}{2} - 12\right) - \left(\frac{12}{2} - 8\right) = \left(\frac{27}{2} - \frac{24}{2}\right) - (6 - 8) = \frac{3}{2} - (-2) = \frac{3}{2} + 2 = \frac{3}{2} + \frac{4}{2} = \frac{7}{2} = 3.5\]

4) \[\int_{1}^{2} 8^x dx = \frac{8^x}{\ln 8} \Big|_{1}^{2} = \frac{8^2}{\ln 8} - \frac{8^1}{\ln 8} = \frac{64}{\ln 8} - \frac{8}{\ln 8} = \frac{56}{\ln 8} = \frac{56}{3 \ln 2} \approx 26.92\]

5) \[\int_{1}^{2} (6x^2 - 8x + 9) dx = \left(2x^3 - 4x^2 + 9x\right) \Big|_{1}^{2} = \left(2(2^3) - 4(2^2) + 9(2)\right) - \left(2(1^3) - 4(1^2) + 9(1)\right) = (16 - 16 + 18) - (2 - 4 + 9) = 18 - 7 = 11\]

6) \[\int_{0}^{2} (2x - 1)^3 dx\] Пусть \[u = 2x - 1, du = 2 dx, dx = \frac{1}{2} du\] Когда \[x = 0, u = 2(0) - 1 = -1\] Когда \[x = 2, u = 2(2) - 1 = 3\] Тогда: \[\frac{1}{2} \int_{-1}^{3} u^3 du = \frac{1}{2} \cdot \frac{u^4}{4} \Big|_{-1}^{3} = \frac{1}{8} \cdot (3^4 - (-1)^4) = \frac{1}{8} \cdot (81 - 1) = \frac{80}{8} = 10\]

7) \[\int_{1}^{3} 3^{4 - 2x} dx\] Пусть \[u = 4 - 2x, du = -2 dx, dx = -\frac{1}{2} du\] Когда \[x = 1, u = 4 - 2(1) = 2\] Когда \[x = 3, u = 4 - 2(3) = -2\] Тогда: \[-\frac{1}{2} \int_{2}^{-2} 3^u du = \frac{1}{2} \int_{-2}^{2} 3^u du = \frac{1}{2} \cdot \frac{3^u}{\ln 3} \Big|_{-2}^{2} = \frac{1}{2 \ln 3} \cdot (3^2 - 3^{-2}) = \frac{1}{2 \ln 3} \cdot \left(9 - \frac{1}{9}\right) = \frac{1}{2 \ln 3} \cdot \frac{80}{9} = \frac{40}{9 \ln 3} \approx 4.05\]