∫上限1下限0e+x+2dx

弓功娅1038∫√xInxdx上限1下限0 求答案 谢了 -
祖狭显17140365671 ______ ∫(上限1下限0)√xInxdx =2/3∫(上限1下限0)Inxdx^(3/2) =2/3*[x^(3/2)lnx(上限1下限0)-∫(上限1下限0)x^(3/2)dlnx] (下限时要求极限limlnx/x^(-3/2)=lim(1/x)/(-3/2)x^(-5/2)=-2/3limx^(3/2)=0(x趋向于0)) =2/3*[0-∫(上限1下限0)x^(1/2)dx] =-2/3*2/3*x^(3/2)(上限1下限0) =-4/9

弓功娅1038∫√(e^x+1)dx 上限ln2下限0 -
祖狭显17140365671 ______ 换元整体令√(e^x+1)=t 所以x=ln(t^2-1) 原式=∫tdln(t^2-1)=∫t*2t/(t^2-1)dt=∫(2t^2-2+2)/(t^2-1)dt=∫[2+2/(t^2-1)]dt=2t|(0,ln2)+∫(1/(t-1)-1/(t+1))dt=2ln2+ln|(t-1)/(t+1)||(0,ln2) 至于你写错了就更简单了 也是令t=√(e^x-1),t=ln(t^2+1) 原式=∫tdln(t^2+1)=∫2t^2/(1+t^2)dt=∫(2t^2+2-2)/(t^2+1)dt=∫2dt-2∫1/(t^2+1)dt=2-2arctant|(0,1)=2-2arctant1=2-π/2

弓功娅1038fx是连续的 f(x)=x/(1+x^2)+(1 - x^2)^(1/2)∫(上限1,下限0)f(x)dx.求∫(上限1,下限0)f(x)dx是学生在考试,求求大家了,不要挂啊!!! -
祖狭显17140365671 ______[答案] a=∫(上限1,下限0)f(x)dx f(x)=x/(1+x^2)+a*(1-x^2)^(1/2) a=∫(上限1,下限0)f(x)dx =∫(上限1,下限0)(x/(1+x^2)+a*(1-x^2)^(1/2))dx =(上限1,下限0)(1/2*ln(x^2+1)+a*(1/4*x*(1-x^2)^(1/2)+1/2*arcsinx)) =1/2*ln2+api/4 a=1/2*ln2+api/4...

弓功娅1038求解定积分11 ∫(上限2,下限0)1+x分之1*dx2 ∫(上限1,下限0)x²+1分之3x的4次方+3x²+5*dx -
祖狭显17140365671 ______[答案] 1 ∫(上限2,下限0)1+x分之1*dx =ln(1+x)|(上限2,下限0) =ln(1+2)-ln(1+0) =ln3-ln1 =ln3 2 ∫(上限1,下限0)x²+1分之3x的4次方+3x²+5*dx =∫(上限1,下限0)[3x²+5/(x²+1)]*dx =[x的立方+5arctanx]|(上限1,下限0) =(1+5派/4)-(0+0) =1+(...

弓功娅1038∫积分上限正无穷下限负无穷 2x/(1+x^2)dx -
祖狭显17140365671 ______[答案] ∫积分上限正无穷下限负无穷 2x/(1+x^2)dx =∫积分上限正无穷下限0 2x/(1+x^2)dx+∫积分上限0,下限负无穷 2x/(1+x^2)dx =ln(1+x²)|(0,+∞)+ln(1+x²)|(-∞,0) 因为 ln(1+(+∞)²)不存在 所以 该积分发散.

弓功娅1038求定积分∫dx/[根号(1+x^2)^3],上限1,下限0. -
祖狭显17140365671 ______ 令x=tant,dx=(sect)^2dt. x=0时t=0,x=1时,t=π/4,所以 ∫(0,1) dx/√[(1+x^2)^3] =∫(0,π/4) cost dt =sin(π/4) =√2/2

弓功娅1038∫dx/(1+x)上限为负2下限为 - e - 1 -
祖狭显17140365671 ______ ∫(上限是1下限是-1)[(x^2+e^x^2)(f(x)-f(-x)]dx=∫(-1,1)(x^2+e^x^2)f(x)dx-∫(-1,1)(x^2+e^x^2)f(-x)dx 对于∫(-1,1)(x^2+e^x^2)f(-x)dx,令-x=t ∫(-1,1)(x^2+e^x^2)f(-x)dx=-∫(1,-1)(t^2+e^t^2)f(t)dt=∫(-1,1)(t^2+e^t^2)f(t)dt=∫(-1,1)(x^2+e^x^2)f(x)dx (x和t地位一样) 所以原式=∫(-1,1)(x^2+e^x^2)f(x)dx-∫(-1,1)(x^2+e^x^2)f(x)dx=0

弓功娅1038计算 ∫上限1下限 - 1 (x^2+2x - 3)dx -
祖狭显17140365671 ______[答案] ∫(从-1到1)(x^2+2x-3)dx=[(x^3/3)+x^2-3x](从-1到1)=1/3+1-3-[(-1/3)+1+3]=1/3+1-3+1/3-1-3=2/3-6=-16/3

弓功娅1038∫(上限1,下限0)x²/x²+1 dx -
祖狭显17140365671 ______ 解: ∫(上限1,下限0) x²/x²+1 dx =∫(上限1,下限0)(x²+1-1)/(x²+1)dx =∫(上限1,下限0)[1-1/(x²+1)]dx =x-arctanx|(上限1,下限0) =(1-arctan1)-(0-arctan0) =(1-π/4)-(0-0) =1-π/4

弓功娅1038∫上限1下限 - 1(x^2*sinx^5+1)/(1+x^2) dx -
祖狭显17140365671 ______ 解∫上限1下限-1 (x^2*sinx^5+1)/(1+x^2) dx=∫上限1下限-1(x^2*sinx^5)/(1+x^2) dx+∫ 上限1下限-1 1/(1+x^2)dx =0+2∫上限1下限0 1/(1+x^2)dx =2arctanx |(上限1下限0) =2(arctan1-arctan0) =π/2 说明:(x^2*sinx^5)/(1+x^2)是奇函数.故它在上限1下限-1的区间内积分为0;1/(1+x^2)为偶函数.故它在上限1下限-1的区间内积分等于:2∫上限1下限0 1/(1+x^2)dx.