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matlab求上限帶有變量的定積分

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情況如圖，我想求出ds的表達式，應(yīng)該是一個含有xs的表達式。并對ds作圖，嘗試過int和quad，都沒有成功

clc;
clear;
A=[129.8665,-55.2435,1.7436,-297.3908;
    52.5130,-5.7480,0.7394,-128.6368;
    59.7245,393.9655,-0.1244,150.0184;
    -7.0125,34.5045,0.0284,-9.5229;
    -4.6130,18.3620,0.0113,-0.3924];
B=[0.2;0.2;225;1];
C=A*B;
a=C(1,1);
b=C(2,1);
smax=C(3,1);
p1=C(4,1);
p2=C(5,1);
syms t;
xs=[-a:a];
ts=2*b./sqrt(1-xs.^2/a^2);
ds=quadl(@(t)fx(t,xs,a,b,smax,p1,p2),0,ts)

fx.m

function f=fx(t,xs,a,b,smax,p1,p2)
f=smax./ts.*(1-xs.^2/a^2).^(p1-1).*(1-(-b*sqrt(1-xs.^2/a^2)+t).^2/(b^2*sqrt(1-xs.^2/a^2))).^(p2-1);
end
回答
編輯回答
乖乖噠

換元 t = ts * z 后，會發(fā)現(xiàn)積分其實是個Beta函數(shù)。

$$ \array{ \text{原式} &=& \fract2vnk71{t_s}(1-\frac{x_s^2}{a^2})^{(\beta_x-1)} \int_{0}^{1} \! \left[1-\frac{(t_s z-t_s/2)^2}{(t_s/2)^2}\right]^{\beta_y-1} t_s\,\mathrmi3pdati z\\ &=& d\, (1-\frac{x_s^2}{a^2})^{(\beta_x-1)} \int_{0}^{1} \! [1-(2z-1)^2]^{\beta_y-1}\,\mathrmelj94mz z\\ &=& 4^{(\beta_y-1)}\, d\, (1-\frac{x_s^2}{a^2})^{(\beta_x-1)}\, \int_{0}^{1} \! z^{\beta_y-1} (1-z)^{\beta_y-1}\,\mathrmj8guqfz z\\ &=& 4^{(\beta_y-1)}\, d\, \mathrm{Beta}(\beta_y, \beta_y)\, (1-\frac{x_s^2}{a^2})^{(\beta_x-1)} } $$

2017年6月20日 08:59