Let
I=∫0α1−tt50dt
⇒I=−∫0α(1−t1−t50−1)dt
⇒I=−∫0α(1−t1−t50)dt+∫0α(1−t1)dt
⇒I=−∫0α(1+t+t2+...+t49)dt−[ln(1−t)]0α
⇒I=−[t+2t2+3t3+...+50t50]0α−ln(1−α)
⇒I=−[α+2α2+3α3+...+50α50]−ln(1−α)
⇒I=−(β+P50(α))
JEE Main 2023 — Mathematics Calculus
Let α∈(0,1) and β=loge(1−α). Let Pn(x)=x+2x2+3x3+….+nxn,x∈(0,1). Then the integral ∫0α1−tt50dt is equal to
Held on 31 Jan 2023 · Verified 6 Jul 2026.
β−P50(α)
−(β+P50(α))
P50(α)−β
β+P50(α)
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