(x+2)dA; R={(x,y):0≤x≤3,1≤y≤4}

Question

Evaluating integrals over region R

Answer 1

Taking dA=dxdy so that dxdy is a tiny rectangle with area dA, and using S[vL,vH](...)dv to denote integral with respect to variable v between lower limit vL and higher limit vH:

S[1,4](S[0,3](x+2)dx)dy=S[1,4](x^2/2+2x)[0,3])dy=S[1,4](9/2+6)dy=S[1,4](21/2)dy=(21/2)(y)[1,4]=(21/2)*3=63/2.