泰勒公式求∫(0→1)(sinx/x)dx求近似值,谁会的教教,

Question

泰勒公式求∫(0→1)(sinx/x)dx求近似值,谁会的教教,

Answer 1

在(0,1)之间sinx的泰勒展开sinx = x - x^3/3!+x^5/5!+...+(-1)^(k-1)*x^(2k-1)/(2k-1)!+...
sinx/x=1-x^2/3!+x^4/5!+...+(-1)^(k-1)x^(2k-2)/(2k-1)!+...
=>
∫(0→1)(sinx/x)dx = (0->1)(x-x^3/(3*3!)+x^5/(5*5!)+...+(-1)^(k-1)x^(2k-1)/((2k-1)*(2k-1)!)+...
取前一项∫(0→1)(sinx/x)dx=1
取前二项∫(0→1)(sinx/x)dx=1-1/18
取前三项∫(0→1)(sinx/x)dx=1-1/18+1/600
...