If B = ∫ 1 e^ x + 1 dx= - f ( x ) + C , where, C is the constant of integration and e^ f(0) = 2, then the value of e^ f(- 1) is

Correct answer: B. e + 1 The given integral is B = ∫ e^ - x 1 + e^ - x dx Let, e^ - x + 1 = t ⇒ - e^ - x dx = dt ∴ B = ∫ (- dt)/(t) = - ln t + C ⇒ B = - ln (e^ - x + 1) + C ∴ f(x) = (e^ - x + 1) Thus, e^ f(- 1) = e^ (e + 1) = e + 1

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If B = ∫ 1 e^ x + 1 dx= - f ( x ) + C , where, C is the constant of...

B = ∫ \dfrac{1}{e^{x} + 1}dx= - f ( x ) + C

, where,

C

is the constant of integration and

e^{f\left(0\right)} = 2,

then the value of

e^{f\left(- 1\right)}

Options

A $4$
B $e + 1$
C $2e$
D $0$

Correct answer

B. $e + 1$

Step-by-step solution

The given integral is

B = ∫ \dfrac{e^{- x}}{1 + e^{- x}}dx

Let,

e^{- x} + 1 = t

⇒ - e^{- x}dx = dt

∴ B = ∫ \dfrac{- dt}{t} = - ln t + C

⇒ B = - ln

\left(e^{- x} + 1\right) + C

<br />

∴ f\left(x\right) = \ln \left(e^{- x} + 1\right)

<br />Thus,

e^{f\left(-

1\right)}

= e^{\ln \left(e + 1\right)} = e + 1

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