∫(1-->3) x/6 dx + ∫(3-->7/2) (2 - x/2) dx
= 1/6 · x¹⁺¹/(1 + 1)|(1-->3) + [2x - 1/2 · x¹⁺¹/(1 + 1)]|(3-->7/2)
= 1/6 · x²/2|(1-->3) + (2x - 1/2 · x²/2)|(3-->7/2)
= 1/12 · (3² - 1²) + [2(7/2) - (1/2)(7/2)²/2] - [2(3) - (1/2)(3²)/2]
= 2/3 + 63/16 - 15/4
= 41/48
∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C
∫(a-->b) f(x) dx = F(b) - F(a),F(x)是f(x)的原函數(shù),即不定積分的結(jié)果那 (2 - x/2) dx怎么得出 [2x - 1/2 · x¹⁺¹/(1 + 1)]有公式嗎這個(gè)∫ x^n dx = [x^(n + 1)]/(n + 1) + C ∫ (2 - x/2) dx = 2∫ 1 dx - (1/2)∫ x dx，常數(shù)可以提取出來 = 2∫ dx - (1/2)∫ x dx = 2x - (1/2)[x^(1 + 1)]/(1 + 1) = 2x - (1/2) · x^2/2 = 2x - (1/4)x^2 其實(shí)常數(shù)的公式∫ kf(x) dx = k · ∫ f(x) dx ∫ k dx = k · ∫ dx = kx + C，被積函數(shù)k是任意常數(shù)