Mathematics › Definite Integration › Definite as Limit of Sum
To find,
limn→∞nn2+12+nn2+22+…+nn2+2n2
=∑r=12nnn2+r2=1n∑r=12n11+rn2
=∫0211+x2dx=tan-1x02
=tan-12.
Asked in: JEE Main 2019 (12 Jan Shift 2)
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