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Question

Mathematics Question on integral

Evaluate the definite integral as limit of sums: ∫abxdx∫^b_a xdx

Answer

The correct answer is: =12(b2−a2)=\frac{1}{2}(b^2-a^2)
It is known that,
∫abƒ(x)dx=(b−a)limn→∞1n[ƒ(a)+ƒ(a+h)+...+ƒ(a+(n−1)h)]∫^b_aƒ(x)dx=(b-a)\underset{n→∞}{lim}\frac{1}{n}[ƒ(a)+ƒ(a+h)+...+ƒ(a+(n-1)h)],where h=b−anh=\frac{b-a}{n}
Here,a=a,b=ba=a,b=b,and ƒ(x)=xƒ(x)=x
∴∫abxdx=(b−a)limn→∞1n[a+(a+h)...(a+2h)...a+(n−1)h]∴∫^b_axdx=(b-a)\underset{n→∞}{lim}\frac{1}{n}[a+(a+h)...(a+2h)...a+(n-1)h]
=(b−a)limn→∞1n[(a+a+a+n times....+a)+(h+2h+3h+...+(n−1)h]=(b-a)\underset{n→∞}{lim}\frac{1}{n}[(a+a+a+n\, times....+a)+(h+2h+3h+...+(n-1)h]
=(b−a)limn→∞1n[na+h(1+2+3+.....+(n−1))]=(b-a)\underset{n→∞}{lim}\frac{1}{n}[na+h(1+2+3+.....+(n-1))]
=(b−a)limn→∞1n[na+h[(n−1)(n)2]]=(b-a)\underset{n→∞}{lim}\frac{1}{n}[na+h[\frac{(n-1)(n)}{2}]]
=(b−a)limn→∞1n[na+[n(n−1)(h)2]]=(b-a)\underset{n→∞}{lim}\frac{1}{n}[na+[\frac{n(n-1)(h)}{2}]]
=(b−a)limn→∞[a+[(n−1)(h)2]]=(b-a)\underset{n→∞}{lim}[a+[\frac{(n-1)(h)}{2}]]
=(b−a)limn→∞[a+[(n−1)(b−a)2n]]=(b-a)\underset{n→∞}{lim}[a+[\frac{(n-1)(b-a)}{2n}]]
=(b−a)limn→∞[a+[(1−1n)(b−a)2]]=(b-a)\underset{n→∞}{lim}[a+[\frac{(1-\frac{1}{n})(b-a)}{2}]]
=(b−a)[a+(b−a)2]=(b-a)[a+\frac{(b-a)}{2}]
=(b−a)[2a+b−a2]=(b-a)[\frac{2a+b-a}{2}]
=(b−a)(b+a)2=\frac{(b-a)(b+a)}{2}
=12(b2−a2)=\frac{1}{2}(b^2-a^2)