f(x)=(sinx+1)e^x在-1到1区间积分,求1到2上的最大值

I =∫(1+sinx)e^xdx =∫(1+sinx)de^x
=[e^x(1+sinx)] - ∫e^xcosxdx
=[e^x(1+sinx)] - ∫e^xcosxdx
= e(1+sin1)-(1-sin1)/e -2∫e^xcosxdx
I1 =∫e^xcosxdx =∫e^xdsinx
=[e^xsinx]-∫e^xsinxdx
=esin1+∫e^xdcosx = esin1+[e^xcosx]-I1,
得 I1 = (esin1+ecos1-1)/2
则 I = e(1+sin1)-(1-sin1)/e -(esin1+ecos1-1)
= -1+e(1+cos1)-(1-sin1)/e.
求谁的最大值? f(x)的?
f'(x)=cosxe^x+(sinx+1)e^x=(1+sinx+cosx)e^x
=[1+√2sin(x+π/4)]e^x=0,
得 sin(x+π/4)=-1/2, 在区间[1,2]上没有驻点.
则最小值是f(1)=(1+sin1)e, 最大值是f(2)=(1+sin2)e^2.