Integral (sin2x + 3cosx) batas atas ⅓π batas bawah 0 dx = 24/07/2022 by rizal ⅓π₀ = (-½cos2(π/3) + 3sinπ/3) – (-½cos0 + 3sin0) = (-½(-½) + 3.½√3) – (-½.1 + 3.0) = 1 ₊ 3√3 ₊ 1 2 2 2 = 3 ₊ 3√3 4 2 = 3 (1 + 2√3) 4 Nilai limit x mendekati -2 x + 6 sin (x + 2)/x² - 3x - 10 =
