计算:不定积分∫1/[(1+2x)(1+x^2)]dx 定积分∫ (x-2) √(4-x^2)dx (上限2,下限-2)
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/08/17 08:07:37
计算:不定积分∫1/[(1+2x)(1+x^2)]dx 定积分∫ (x-2) √(4-x^2)dx (上限2,下限-2)
![计算:不定积分∫1/[(1+2x)(1+x^2)]dx 定积分∫ (x-2) √(4-x^2)dx (上限2,下限-2)](/uploads/image/z/1552893-69-3.jpg?t=%E8%AE%A1%E7%AE%97%EF%BC%9A%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB1%2F%5B%281%2B2x%EF%BC%89%281%2Bx%5E2%29%5Ddx+%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB+%28x-2%29+%E2%88%9A%284-x%5E2%29dx+%EF%BC%88%E4%B8%8A%E9%99%902%2C%E4%B8%8B%E9%99%90-2%EF%BC%89)
待定系数法:
1/[(1 + 2x)(1 + x^2)] = a/(1 + 2x) + (bx + c)/(1 + x^2)
1 = a(1 + x^2) + (bx + c)(1 + 2x)
当x = -1/2,1 = 1.25a => a = 4/5
当x = 0,1 = 4/5 + c => c = 1/5
当x = 1,1 = 8/5 + (b + 1/5)(3) => b = -2/5
∫ dx/[(1 + 2x)(1 + x^2)]
= (4/5)∫ dx/(1 + 2x) - (2/5)∫ x/(1 + x^2) dx + (1/5)∫ dx/(1 + x^2)
= (4/5)(1/2)ln|1 + 2x| - (2/5)(1/2)ln(1 + x^2) + (1/5)arctan(x) + C
= (2/5)ln|1 + 2x| - (1/5)ln(1 + x^2) + (1/5)arctan(x) + C
__________________________________________________________
∫(-2到2) (x - 2)√(4 - x^2) dx
= ∫(-2到2) x√(4 - x^2) dx - 2∫(-2到2) √(4 - x^2) dx
= 0 - 4∫(0到2) √(4 - x^2) dx
用几何意义解比较快速
表示的圆是x^2 + y^2 = 4,半径为2,
范围由-2到2表示半圆面积
范围由0到2表示1/4的圆面积
= π(2)² * 1/4 = π
所以定积分 = -4(π) = -4π
方法二:用第二类换元法,用代换x = 2sinθ即可.
1/[(1 + 2x)(1 + x^2)] = a/(1 + 2x) + (bx + c)/(1 + x^2)
1 = a(1 + x^2) + (bx + c)(1 + 2x)
当x = -1/2,1 = 1.25a => a = 4/5
当x = 0,1 = 4/5 + c => c = 1/5
当x = 1,1 = 8/5 + (b + 1/5)(3) => b = -2/5
∫ dx/[(1 + 2x)(1 + x^2)]
= (4/5)∫ dx/(1 + 2x) - (2/5)∫ x/(1 + x^2) dx + (1/5)∫ dx/(1 + x^2)
= (4/5)(1/2)ln|1 + 2x| - (2/5)(1/2)ln(1 + x^2) + (1/5)arctan(x) + C
= (2/5)ln|1 + 2x| - (1/5)ln(1 + x^2) + (1/5)arctan(x) + C
__________________________________________________________
∫(-2到2) (x - 2)√(4 - x^2) dx
= ∫(-2到2) x√(4 - x^2) dx - 2∫(-2到2) √(4 - x^2) dx
= 0 - 4∫(0到2) √(4 - x^2) dx
用几何意义解比较快速
表示的圆是x^2 + y^2 = 4,半径为2,
范围由-2到2表示半圆面积
范围由0到2表示1/4的圆面积
= π(2)² * 1/4 = π
所以定积分 = -4(π) = -4π
方法二:用第二类换元法,用代换x = 2sinθ即可.
计算:不定积分∫1/[(1+2x)(1+x^2)]dx 定积分∫ (x-2) √(4-x^2)dx (上限2,下限-2)
计算定积分 ∫ x ln(1+e^x) dx (上限2下限-2)
计算定积分∫1/x(x+1) dx (上限2 下限 1)
求定积分∫|x|dx,上限1,下限-2
计算定积分 ∫(x+3)/根号(2x+1)dx,上限4,下限0
用定义计算定积分 ∫上限1 下限-1 (x^2+1-x)dx
求定积分:∫dx/x(根号x^2-1),上限 - (根号2),下限-2
求解定积分∫(上限1,下限0)ln(x+1)/(2-x)^2.dx
计算定积分 ∫1 0 (e^x-e^-x)^2dx 上限1 下限0
计算定积分∫ x(1-2x)^2 √1-x^2 dx 下限-1,上限1
计算定积分∫(上限1下限-1) (2x^2+x^3*cosx)/(1+√(1-x^2))dx
求计算定积分ln(x+√(x^2+1))dx ,上限1,下限0