Здравствуйте, Аня Васильева.

1. Первый способ:
∫(5 – 2x³)x²dx = -1/6 ∙ ∫(5 – 2x³)d(5 – 2x³) = -1/6 ∙ 1/2 ∙ (5 – 2x³)² + C = -1/12 ∙ (5 – 2x³)² + C,
₀∫¹(5 – 2x³)x²dx = -1/12 ∙ (5 – 2x³)²|₀¹ = -1/12 ∙ (3² – 5²) = 16/12 = 4/3 ≈ 1,33.
Второй способ:
∫(5 – 2x³)x²dx = 5∫x²dx – 2∫x⁵dx = 5/3 ∙ x³ – 1/3 ∙ x⁶ + C,
₀∫¹(5 – 2x³)x²dx = 5/3 ∙ x³|₀¹ – 1/3 ∙ x⁶|₀¹ = 5/3 – 1/3 = 4/3 ≈ 1,33.

2. ∫e^{sin x}cos xdx = ∫e^{sin x}d(sin x) = e^{sin x} + C,
₀∫⁶e^{sin x}cos xdx = e^{sin x}|₀⁶ = e^{sin 6} – e^{sin 0} = e^{sin 6} – 1 ≈ -0,244.

С уважением.