If $\mathop \smallint \nolimits_0^1 \,\,\frac{{{e^t}\,dt}}{{t + 1}} = a,$, then $\mathop \smallint \nolimits_{b - 1}^b \,\,\frac{{{e^{ - t}}\,dt}}{{t - b - 1}}$ is equal to

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Analyzing +4 | -1

\int \nolimits_0^1 \,\,\frac{{{e^t}\,dt}}{{t + 1}} = a,

, then

\int \nolimits_{b - 1}^b \,\,\frac{{{e^{ - t}}\,dt}}{{t - b - 1}}

is equal to

View answer and explanation

Correct option: B

- \,a{e^{ - b}}

Option B (

-ae^{-b}

) is correct because after substitution and manipulation of the integral, it simplifies to

-ae^{-b}

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