高数定积分题目及答案 定积分100道例题及解答

y'=根号下cosx,于是弧微分ds=根号下{1+(根号下cosx)^2}dx=根号下(1+cosx)dx.注意到x从-pai/2变到pai/2曲线就获得了全长,所求曲线长是 s=定积分(从-pai/2到pai/2)根号下(1+cosx)dx=定积分(从-pai/2到pai/2)根号下[2cos^2 (x/2)]dx=根号下2 * 定积分(从-pai/2到pai/2)cos (x/2)]dx=2(根号下2)* sin(x/2){上pai/2、下-pai/2}=2(根号下2)*(根号下2)=4.

分段求积分=∫(1,2) x/√(4-x^2)dx+∫(2,3) x/√(x^2-4)dx=-√(4-x^2)|(1,2) +√(x^2-4)|(2,3) =√3+√5

(1) ∫(0->π/2) xsin2x dx=-(1/2)∫(0->π/2) x dcos2x=-(1/2)[ xcos2x]|(0->π/2) +(1/2)∫(0->π/2) cos2x dx=-π/2+ (1/2)∫(0->π/2) cos2x dx=-π/2 +(1/4)[sin2x]|(0->π/2)=-π/2(2) ∫(0->+∞) x.e^(.

解:令√x=sint 原式=∫(上限π/6,下限0) (1 + 2sint - cos2t)dt=(24 + 2π - 15√3)/12.

原式=(-1/3)*∫ arcsinx d[(1-x^2)^(3/2)]=(-1/3)*[arcsinx*(1-x^2)^(3/2)-∫ (1-x^2)^(3/2) d(arcsinx)]=(-1/3)*[arcsinx*(1-x^2)^(3/2)-∫ (1-x^2)^(3/2)*(1-x^2)^(-1/2) dx]=(-1/3)*[arcsinx*(1-x^2)^(3/2)-∫ (1-x^2) dx]=(-1/3)*[arcsinx*(1-x^2)^(3/2)-x+x^3/3]+C=x/3-(1/3)*arcsinx*(1-x^2)^(3/2)-(1/9)*x^3+C

sin^2(x)在π/4到5π/4范围内最大值是1,最小值是0,所以1+0《1+sin^2(x)《1+1

∫(0->1) f''(x/2) dx^2=2∫(0->1) x.f''(x/2) dx=4∫(0->1) x.df'(x/2) =4[x.df'(x/2)]|(0->1) -4∫(0->1) f'(x/2) dx=4f'(1/2) -8[f(x/2)]|(0->1)=16- 8[f(1/2) -f(0)]=16- 8(2-1)=16-8=8

<p></p> <p> </p> <p>前面的人把第二题算错了.</p> <p> </p> <p>【数学之美】团队为您解答,若有不懂请追问,如果解决问题请点下面的“选为满意答案”.</p>

解:原式=[(x^2/2)ln((1+x)/(1-x))]│-∫x^2dx/(1-x^2) (应用分部积分法) =ln3/8-[x-(1/2)ln((1+x)/(1-x))]│ =ln3/8-(1-ln3)/2 =5ln3/8-1/2.

如果方便,你可以直接到百度文库上搜索,上面有很多类似资料. 令建议你将定积分和不定积分分开来搜索.