众文网1.matlab求解带参数的定积分,毕业设计急用,编程大神来帮忙啊啊啊,
>> t=0:0.1:0.5; >> arrayfun(@(t)quad(@(x)x.*(t-x).^2.*sqrt(1.1*10^21*x.^2-7.3),0,t),t)ans = 1.0e+007 * 0 0.001105541596785 0.035377331097124 0.268646608018787 1.132074595107977 3.454817489953542建议你学习两个函数的用法即可数值积分函数 quad（)，如果你是新版matlab，好像是用integral（)函数，当然quad（)也还能用对输入参数进行数组化计算arrayfun()。
2.定积分求解
∫[-π/2，π/2] (sinx^2)dx/(1+e^x)
=(1/2)∫[-π/2，π/2] (1-cos2x)dx/(1+e^x)
=(1/2)∫[-π/2，π/2]d-e^(-x)/(1+e^(-x)+(-1/2)∫[-π/2， π/2] cos2xdx/(1+e^x)
=(1/2)( -ln(1+e^(-x)） ）|[-π/2， π/2] +(-1/2)∫[-π/2， π/2]cos2xdx/(1+e^x)
=(1/2)ln(1+e^（-π/2）)/(1+e^（π/2）)] -(1/2)∫[-π/2，π/2] cos2xdx/(1+e^x)
=(1/2)ln|(1+e^（-π/2）)/(1+e^（π/2）)|
∫[-π/2，π/2] cos2xdx/(1+e^x)
=(-1/2)∫[-π/2， π/2]dsin2x/(1+e^x)
=-(1/2)sin2x/(1+e^x)| [-π/2， π/2] (-1/2)∫[-π/2，π/2]sin2xe^xdx/(1+e^x)^2
=(-1/2)∫[-π/2,0] sin2xe^xdx/(1+e^x)^2+(-1/2)∫[0，π/2] sin2xe^xdx/(1+e^x)^2
=(1/2)∫[0，-π/2]sin2xe^xdx/(1+e^x)^2 +(-1/2)∫[0，-π/2] sin2ue^(-u)du/(1+e^(-u))^2 （其中u=-x）
=(1/2)∫[0，-π/2]sin2xe^xdx/(1+e^x)^2+(-1/2)∫[0， -π/2]sin2ue^udu/(1+e^u)^2
=0
f(x)=sin2xe^x/(1+e^x)^2
f(-x)=sin(-2x)e^(-x)/(1+e^(-x))^2=-sin2xe^x/(1+e^x)^2=-f(x)
奇函数的对称积分为0 如：∫[-a,a] sinxdx=-cosa+cos(-a)=0
3.定积分求解
∫【-a,a】f(x)dx，若f(x)是奇函数，值为零。
∫【-2,2】[x³cos(x/2)+1/2]√（4-x²）dx
=∫【-2,2】0.5√（4-x²）dx【x=2siny】
=∫【-π/2，π/2】√（1-sin²y）d(2siny)
=∫【-π/2，π/2】2cos²ydy
=∫【-π/2，π/2】[1+cos(2y)dy
=y+0.5sin(2y)【上限π/2，下限-π/2】
=π
4.求解定积分
解：享种解
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=sinydt=cosydyy∈[0，π/2]
∴∫（0,1）(t^4)(1-t)dt/√（1-t^2）=∫（0，π/2）(1-siny)(siny)^4dy=∫（0，π/2）[(siny)^4-(siny)^5]dy
(siny)^4=(1/4)(1-cos2y)^2=(1/8)(3-4cos2y+cos4y)、（siny）^5dy=-[1-(cosy)^2]^2d(cosy)=-[1-(cosy)^2]^2d(cosy)=-[1-2(cosy)^2+(cosy)^4]d(cosy)
∴∫（0,1）(t^4)(1-t)dt/√（1-t^2）={(1/8)[3y-2sin2y+(1/4)sin4y]+[cosy-(2/3)(cosy)^3+(1/5)(cosy)^5]}丨（y=0，π/2）=3π/16-8/15
∴原式=50(3π/8-16/15)=5(45π/8-128)/12
供参考
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