Let function F be defined as f ( x ) = _1^x e^t t dt ,, ,x > 0 th… - Vidyadip

MCQ

Let function $F$ be defined as $f\left( x \right) = \int\limits_1^x {\frac{{{e^t}}}{t}dt\,,\,x > 0} $ then the value of the integral $\int\limits_1^x {\frac{{{e^t}}}{{t + a}}dt\,} $ , where $a > 0$ , is

A
${e^a}\left[ {F\left( x \right) - F\left( {1 + a} \right)} \right]$
B
${e^{ - a}}\left[ {F\left( {x + a} \right) - F\left( a \right)} \right]$
C
${e^a}\left[ {F\left( {x + a} \right) - F\left( 1+a \right)} \right]$
✓
${e^{ - a}}\left[ {F\left( {x + a} \right) - F\left( {1 + a} \right)} \right]$

✓

Answer

Correct option: D.

${e^{ - a}}\left[ {F\left( {x + a} \right) - F\left( {1 + a} \right)} \right]$

d
$F(x)=\int_{1}^{x} \frac{e^{t}}{t} d t, x>0$

Let $I=\int_{1}^{x} \frac{e^{t}}{t+a} d t$

Put $t+a=z \Rightarrow t=z-a ; d t=d z$

for $t=1, z=1+a$

for $t=x, z=x+a$

$\therefore \mathrm{I}=\int_{1+a}^{x+a} \frac{e^{z-a}}{z} d t$

$=e^{-a \int_{1+a}^{x+a}} \frac{e^{z}}{z} d z=e^{-a} \int_{1+a}^{x+a} \frac{e^{d}}{t} d t$

$\mathrm{I}=e^{-a}\left[\int_{1+a}^{1} \frac{e^{t}}{t} d t+\int_{1}^{x+a} \frac{e^{t}}{t} d t\right]$

$=e^{-a}\left[-\int_{1}^{1+a} \frac{e^{t}}{t} d t+\int_{1}^{x+a} \frac{e^{t}}{t} d t\right]$

$=e^{-a}[-F(1+a)+F(x+a)]$

(By the definition of $\mathrm{F}(\mathrm{x})$ )

$=e^{-a}[F(x+a)-F(1+a)]$

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