Question

估计定积分的值：∫<1,2> X/(1+X^2) dX

题目上说是要用矩形法或者梯形法或者抛物线法进行估算0.0 各位大神帮帮再帮帮我吧 感激不尽

Answer 1

∫<1,2> X/(1+X^2) dX=1/2∫<1,2> 1/(1+X^2) dX^2=[ln(1+X^2)<1,2> ]=(ln5-ln2)/2

∫XdX=X^2/2

∫ dX^2=X^2=2∫XdX

Answer 2

1/2*LN(X^2+1)|(2,1)=(ln5-ln2)/2
等于0.4581453659

Answer 3

1/2*LN(X^2+1)|(2,1)=(ln5-ln2)/2