Вычислите следующие определенные интегралы 5. 1) ∫02 x² dx; 2) ∫12 x³ dx; 3) ∫13 x⁴ dx. 6. 1) ∫-23 (4x³-3x²+2x+1)dx; 2) ∫-10 (x³+2x) dx. 7. 1) ∫1/21 dx/x³; 2) ∫1/31/2 dx/x²; 3) ∫04 √x dx. 8. 1) ∫18 ³√x² dx; 2) ∫827 dx/³√x; 3) ∫14 (√x-1/√x) dx; 4) ∫19 (x-1)/√x dx.

Задача 5

1) ∫₀² x² dx

\[\int_{0}^{2} x^2 dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}\]

2) ∫₁² x³ dx

\[\int_{1}^{2} x^3 dx = \left[ \frac{x^4}{4} \right]_1^2 = \frac{2^4}{4} - \frac{1^4}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}\]

3) ∫₁³ x⁴ dx

\[\int_{1}^{3} x^4 dx = \left[ \frac{x^5}{5} \right]_1^3 = \frac{3^5}{5} - \frac{1^5}{5} = \frac{243}{5} - \frac{1}{5} = \frac{242}{5}\]

Задача 6

1) ∫₋₂³ (4x³-3x²+2x+1) dx

\[\int_{-2}^{3} (4x^3 - 3x^2 + 2x + 1) dx = \left[ x^4 - x^3 + x^2 + x \right]_{-2}^{3} = (3^4 - 3^3 + 3^2 + 3) - ((-2)^4 - (-2)^3 + (-2)^2 + (-2)) = (81 - 27 + 9 + 3) - (16 + 8 + 4 - 2) = 66 - 26 = 40\]

2) ∫₋₁⁰ (x³+2x) dx

\[\int_{-1}^{0} (x^3 + 2x) dx = \left[ \frac{x^4}{4} + x^2 \right]_{-1}^{0} = (0) - (\frac{(-1)^4}{4} + (-1)^2) = -(\frac{1}{4} + 1) = -\frac{5}{4}\]

Задача 7

1) ∫¹/₂¹ dx/x³

\[\int_{1/2}^{1} \frac{dx}{x^3} = \int_{1/2}^{1} x^{-3} dx = \left[ \frac{x^{-2}}{-2} \right]_{1/2}^{1} = \left[ -\frac{1}{2x^2} \right]_{1/2}^{1} = -\frac{1}{2(1)^2} - \left(-\frac{1}{2(1/2)^2}\right) = -\frac{1}{2} + \frac{1}{2(1/4)} = -\frac{1}{2} + \frac{1}{1/2} = -\frac{1}{2} + 2 = \frac{3}{2}\]

2) ∫¹/₃¹/₂ dx/x²

\[\int_{1/3}^{1/2} \frac{dx}{x^2} = \int_{1/3}^{1/2} x^{-2} dx = \left[ \frac{x^{-1}}{-1} \right]_{1/3}^{1/2} = \left[ -\frac{1}{x} \right]_{1/3}^{1/2} = -\frac{1}{1/2} - \left(-\frac{1}{1/3}\right) = -2 + 3 = 1\]

3) ∫₀⁴ √x dx

\[\int_{0}^{4} \sqrt{x} dx = \int_{0}^{4} x^{1/2} dx = \left[ \frac{x^{3/2}}{3/2} \right]_0^4 = \left[ \frac{2}{3}x^{3/2} \right]_0^4 = \frac{2}{3}(4^{3/2}) - \frac{2}{3}(0^{3/2}) = \frac{2}{3}(8) - 0 = \frac{16}{3}\]

Задача 8

1) ∫₁⁸ ³√x² dx

\[\int_{1}^{8} \sqrt[3]{x^2} dx = \int_{1}^{8} x^{2/3} dx = \left[ \frac{x^{5/3}}{5/3} \right]_1^8 = \left[ \frac{3}{5}x^{5/3} \right]_1^8 = \frac{3}{5}(8^{5/3}) - \frac{3}{5}(1^{5/3}) = \frac{3}{5}(32) - \frac{3}{5}(1) = \frac{96}{5} - \frac{3}{5} = \frac{93}{5}\]

2) ∫₈²⁷ dx/³√x

\[\int_{8}^{27} \frac{dx}{\sqrt[3]{x}} = \int_{8}^{27} x^{-1/3} dx = \left[ \frac{x^{2/3}}{2/3} \right]_8^{27} = \left[ \frac{3}{2}x^{2/3} \right]_8^{27} = \frac{3}{2}(27^{2/3}) - \frac{3}{2}(8^{2/3}) = \frac{3}{2}(9) - \frac{3}{2}(4) = \frac{27}{2} - \frac{12}{2} = \frac{15}{2}\]

3) ∫₁⁴ (√x-1/√x) dx

\[\int_{1}^{4} (\sqrt{x} - \frac{1}{\sqrt{x}}) dx = \int_{1}^{4} (x^{1/2} - x^{-1/2}) dx = \left[ \frac{x^{3/2}}{3/2} - \frac{x^{1/2}}{1/2} \right]_1^4 = \left[ \frac{2}{3}x^{3/2} - 2x^{1/2} \right]_1^4 = (\frac{2}{3}(4^{3/2}) - 2(4^{1/2})) - (\frac{2}{3}(1^{3/2}) - 2(1^{1/2})) = (\frac{2}{3}(8) - 2(2)) - (\frac{2}{3} - 2) = (\frac{16}{3} - 4) - (\frac{2}{3} - 2) = \frac{16}{3} - 4 - \frac{2}{3} + 2 = \frac{14}{3} - 2 = \frac{8}{3}\]

4) ∫₁⁹ (x-1)/√x dx

\[\int_{1}^{9} \frac{x-1}{\sqrt{x}} dx = \int_{1}^{9} (\frac{x}{\sqrt{x}} - \frac{1}{\sqrt{x}}) dx = \int_{1}^{9} (x^{1/2} - x^{-1/2}) dx = \left[ \frac{x^{3/2}}{3/2} - \frac{x^{1/2}}{1/2} \right]_1^9 = \left[ \frac{2}{3}x^{3/2} - 2x^{1/2} \right]_1^9 = (\frac{2}{3}(9^{3/2}) - 2(9^{1/2})) - (\frac{2}{3}(1^{3/2}) - 2(1^{1/2})) = (\frac{2}{3}(27) - 2(3)) - (\frac{2}{3} - 2) = (18 - 6) - (\frac{2}{3} - 2) = 12 - \frac{2}{3} + 2 = 14 - \frac{2}{3} = \frac{40}{3}\]