(a) A=0∫1(1+x+x2)dx=(x+2x2+3x3)∣∣01=(1+21+31)−0=165=611.
(b) An=i=1∑n2f(ni−1)+f(ni)⋅n1=n1(2f(0)+f(n1)+…+f(nn−1)+2f(1))=n1(21+i=1∑n−1f(ni)+23)=n1(i=1∑n−1n2n2+in+i2+2)=n1(n2n2(n−1)+n(n(n−1)/2)+(n−1)n(2n−1)/6+23)=n1(6n11n2−12n+1+23)=n1(6n11n2+1)=611+6n21.
(c) When n→∞, 6n21→0. Therefore, An→611.
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