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Question

Evaluate the following definite integrals as limit of sums.
32x2dx.

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Solution

We know that
baf(x)=limh0h[f(a)+f(a+h)+f(a+2h)+...+f{a+(n1)h}]
where, nh=b -a
Here and a=2, b=3 and nh=b -a=3-2=1
f(x)=x2,f(2)=22,f(2+h)=(2+h)2f(2+2h)=(2+2h)2,.....,f(2+(n1)h)=(2+(n1)h)232x2dx=limh0h[f(2)+f(2+h)+f(2+2h)+....+f(2+(n1)h)]
=limh0h[22+(2+h)2+(2+2h)2+...+(2+(n1)h)2]=limh0h[22+(22+h2+4h)+(22+22h2+8h)+...+(22+(n1)2h2+4(n1)h)][(a+b)2=a2+b2+2ab]=limh0h[(22+22+...+n times)+h2(12+22+32)+....+(n1)2+4h(1+2+...+(n1))]=limh0{4nh+h3(n1)n(2n1)6+4h2(n1)n2}n2=n(n+1)(2n+1)6(n1)2=16(n1)n(2n1)and(n1)=n(n1)2

=limh0{4nh+(hnh)hn(2nhh)6+2(nhh)(nh)}=limh0{4×1+(1h)1(2×1h)6+2(1h)1}[nh=1]=limh0[4+(1h)(2h)6+2(1h)]=4+26+2=6+13=193


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