First slide

Evaluation of definite integrals

Question

If $\int_{0}^{1} \frac{e^{t} dt}{t + 1} = a$ then, $\int_{b - 1}^{b} \frac{e^{- t} dt}{t - b - 1}$ is equal to

Moderate

A

${ae}^{- b}$
B

$- {ae}^{- b}$
C

$- {be}^{- a}$
D

${ae}^{b}$

Solution

$\begin{aligned} \int_{b - 1}^{b} \frac{e^{- t}}{t - b - 1} dt = \int_{- 1}^{0} \frac{e^{- (t + b)}}{(t + b) - b - 1} dt \\ = e^{- b} \int_{- 1}^{0} \frac{e^{- t}}{t - 1} dt = e^{- b} \int_{0}^{1} \frac{e^{- (t - 1)}}{t - 2} dt \end{aligned}$
Put,
$\begin{array}{l} t - 1 = - s \Rightarrow dt = - ds \\ = - e^{- b} \int_{1}^{0} \frac{e^{s}}{- (s + 1)} ds = e^{- b} \int_{1}^{0} \frac{e^{s}}{s + 1} ds \\ = - e^{- b} \int_{0}^{1} \frac{e^{t}}{t + 1} dt = - {ae}^{- b} \end{array}$

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