Application of Integrals

∫₀^π sin x dx, ∫₀^π |sin x| dx, 2∫₀^(π/2) sin x dx

Since sin x ≥ 0 on [0,π], the area is ∫₀^π sin x dx = 2. The absolute value notation is also correct. Due to symmetry, 2∫₀^(π/2) sin x dx also gives the same result.

True

∫₋₁¹ x³ dx = [x⁴/4]₋₁¹ = 1/4 - 1/4 = 0. This is because x³ is an odd function, so the areas above and below the x-axis cancel out over symmetric intervals.