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on solving i found that A(x) is a parabola and the coefficientof x2 is positive so it will open upwardand its local minimum exist for that value x where itsderivative is equal to = 01 Solution{:[area=A(x)=int_(0)^(x)(1+t)dt],[A(x)=[t+(t^(2))/(2)]_(0)^(x)],[A(x)=x+(x^(2))/(2)]:}(a){:[:.A(0)=0+0=0],[A(1)=1+(1)/(2)=(3)/(2)=1.5],[A(2)=2+(2^(2))/(2)=4 ... See the full answer