取F(x)=-cosx+3x^2/2
则F'(x)=sinx+3x
∴∫(π/2,0)(sinx+3x)dx
=∫(π/2,0)F'(x)dx
=F(π/2)-F(0)
=-cosπ/2+3(π/2)^2/2-(-cos0+3*0/2)
=3π^2/8+1