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Definite Integration question

2023 · 31 Jan · Shift 1 · Q33
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  5. /2023 · 31 Jan · Shift 1 · Q33

Definite Integration question

2023 · 31 Jan · Shift 1 · Q33

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let α∈(0,1)\alpha \in (0,1)α∈(0,1) and β=log⁡e(1−α)\beta = {\log _e}(1 - \alpha )β=loge​(1−α). Let Pn(x)=x+x22+x33 + ... + xnn,x∈(0,1){P_n}(x) = x + {{{x^2}} \over 2} + {{{x^3}} \over 3}\, + \,...\, + \,{{{x^n}} \over n},x \in (0,1)Pn​(x)=x+2x2​+3x3​+...+nxn​,x∈(0,1). Then the integral ∫0αt501−tdt\int\limits_0^\alpha {{{{t^{50}}} \over {1 - t}}dt}0∫α​1−tt50​dt is equal to
  1. A
    −(β+P50(α))- \left( {\beta + {P_{50}}\left( \alpha \right)} \right)−(β+P50​(α))
  2. B
    β−P50(α)\beta - {P_{50}}(\alpha )β−P50​(α)
  3. C
    P50(α)−β{P_{50}}(\alpha ) - \betaP50​(α)−β
  4. D
    β+P50−(α)\beta + {P_{50}} - (\alpha )β+P50​−(α)
View written solutionFree

Correct answer: A

  1. Given integral

We need to evaluate I=∫0αt501−t dt,I=\int_0^\alpha \frac{t^{50}}{1-t}\,dt,I=∫0α​1−tt50​dt, where α∈(0,1)\alpha\in(0,1)α∈(0,1) and β=ln⁡(1−α),\beta=\ln(1-\alpha),β=ln(1−α), Pn(x)=x+x22+x33+⋯+xnn.P_n(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots+\frac{x^n}{n}.Pn​(x)=x+2x2​+3x3​+⋯+nxn​.

  1. Rewrite the integrand

Use the identity 11−t=1+t+t2+⋯+t49+t501−t\frac{1}{1-t}=1+t+t^2+\cdots+t^{49}+\frac{t^{50}}{1-t}1−t1​=1+t+t2+⋯+t49+1−tt50​ which implies t501−t=11−t−(1+t+t2+⋯+t49).\frac{t^{50}}{1-t}=\frac{1}{1-t}-\left(1+t+t^2+\cdots+t^{49}\right).1−tt50​=1−t1​−(1+t+t2+⋯+t49).

So, I=∫0α11−t dt−∫0α(1+t+t2+⋯+t49) dt.I=\int_0^\alpha \frac{1}{1-t}\,dt-\int_0^\alpha (1+t+t^2+\cdots+t^{49})\,dt.I=∫0α​1−t1​dt−∫0α​(1+t+t2+⋯+t49)dt.

  1. Integrate term by term

First,

\left[-\ln(1-t)\right]_0^\alpha=-\ln(1-\alpha).$$ Since $\beta=\ln(1-\alpha)$, $$-\ln(1-\alpha)=-\beta.$$ Next, $$\int_0^\alpha (1+t+t^2+\cdots+t^{49})\,dt =\alpha+\frac{\alpha^2}{2}+\frac{\alpha^3}{3}+\cdots+\frac{\alpha^{50}}{50} =P_{50}(\alpha).$$ Hence, $$I=-\beta-P_{50}(\alpha)=-(\beta+P_{50}(\alpha)).$$ 4. **Match with options** This is exactly **Option A**: $$-\left(\beta+P_{50}(\alpha)\right).$$ 5. **Verification with stored answer** Stored correct answer: **A** Our derived answer: **A** So they agree.
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