定积分计算题 定积分练习题
\u5b9a\u79ef\u5206\u8ba1\u7b97\u9898\u7b2c\u4e00\u9053\u79ef\u5206\u9898\u7684\u7ed3\u679c\u4e3a\uff1a1/3\uff0c\u7b2c\u4e8c\u9053\u79ef\u5206\u7ed3\u679c\u7684\u4e3a\uff1a\u03c0/6\u3002
\u8ba1\u7b97\u8fc7\u7a0b\uff1a
1\u3001\u222b\uff080\uff0c1\uff09\u221ax/2dx
=1/2\u222b\uff080\uff0c1\uff09\u221axdx
=\uff081/2\uff09*\uff082/3\uff09*x^\uff083/2\uff09|\uff080\uff0c1\uff09
=1/2*\uff082/3\uff09
=1/3
2\u3001\u03c0\u222b\uff080\uff0c1\uff09x/4dx
=\u03c0\uff08x*x/8\uff09|\uff080\uff0c1\uff09
=\u03c0/8
\u6269\u5c55\u8d44\u6599\uff1a
\u542b\u221a\uff08a+bx\uff09\u7684\u79ef\u5206\uff1a
\u542b\u6709\u221a\uff08a+bx\uff09\u7684\u79ef\u5206\u516c\u5f0f\u4e3b\u8981\u5305\u542b\u6709\u4ee5\u4e0b\u51e0\u7c7b\uff1a
1\u3001
2\u3001
3\u3001
4\u3001
5\u3001
6\u3001
\u5b9a\u79ef\u5206\u4e00\u822c\u5b9a\u7406\uff1a
\u5b9a\u74061\uff1a\u8bbef(x)\u5728\u533a\u95f4[a\uff0cb]\u4e0a\u8fde\u7eed\uff0c\u5219f(x)\u5728[a\uff0cb]\u4e0a\u53ef\u79ef\u3002
\u5b9a\u74062\uff1a\u8bbef(x)\u533a\u95f4[a\uff0cb]\u4e0a\u6709\u754c\uff0c\u4e14\u53ea\u6709\u6709\u9650\u4e2a\u95f4\u65ad\u70b9\uff0c\u5219f(x)\u5728[a\uff0cb]\u4e0a\u53ef\u79ef\u3002
\u5b9a\u74063\uff1a\u8bbef(x)\u5728\u533a\u95f4[a\uff0cb]\u4e0a\u5355\u8c03\uff0c\u5219f(x)\u5728[a\uff0cb]\u4e0a\u53ef\u79ef\u3002
\u5982\u679c\u5728\u533a\u95f4[a,b]\u4e0a\uff0cf(x)\u22650,\u5219\u222b\uff08a\uff0cb\uff09f\uff08x\uff09dx>=0\u3002
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-\u79ef\u5206\u516c\u5f0f
\u5982\u56fe\u6240\u793a
①设x=-t，∴原式=100∫(0,1)(t^4)√[(1-t)/(1+t)]dt=100∫(0,1)(t^4)(1-t)dt/√(1-t^2)。
②设t^2=y，dt=(1/2)dy/√y，
∴∫(0,1)(t^4)(1-t)dt/√(1-t^2)=(1/2)∫(0,1)[y^(3/2)-y^2]dy/√(1-y)=(1/2)[B(5/2,1/2)-B(3,1/2)]，
而利用贝塔函数和伽玛函数的关系式，及伽玛函数的递推式，有B(5/2,1/2)-B(3,1/2)=Γ(5/2)Γ(1/2)/Γ(3)-Γ(3)Γ(1/2)/Γ(7/2)=3π/8-16/15，
∴原式=50(3π/8-16/15)=5(45π/8-128)/12。
供参考。
（-8/15+3*pi/16）*100
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