Mathematics › Definite Integration › Definite as Limit of Sum
Given, a=limn→∞∑k=1n2nn2+k2⇒a=1n∑k=1n21+kn2=∫0121+x2dx⇒a=2tan-1x01⇒a=2π4-0=π2Now, fx=1-cosx1+cosx⇒fx=tanx2;x∈0,1So, fa2=fπ4⇒fπ4=2-1And f'π4=12sec2π8=22+1So, f'π4=2fπ4
Given, a=limn→∞∑k=1n2nn2+k2
⇒a=1n∑k=1n21+kn2=∫0121+x2dx
⇒a=2tan-1x01
⇒a=2π4-0=π2
Now, fx=1-cosx1+cosx
⇒fx=tanx2;x∈0,1
So, fa2=fπ4
⇒fπ4=2-1
And f'π4=12sec2π8=22+1
So, f'π4=2fπ4
Asked in: JEE Main 2022 (26 Jul Shift 1)
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