A) \[{{e}^{a}}\left[ F(x)-F(1+a) \right]\]
B) \[{{e}^{-a}}\left[ F(x+a)-F(a) \right]\]
C) \[{{e}^{a}}\left[ F(x+a)-F(1+a) \right]\]
D) \[{{e}^{-a}}\left[ F(x+a)-F(1+a) \right]\]
Correct Answer: D
Solution :
\[F(x)=\int_{1}^{x}{\frac{{{e}^{t}}}{t}}dt,x>0\] Let\[I=\int_{1}^{x}{\frac{{{e}^{t}}}{t+a}}dt\] Put\[t+a=z\Rightarrow t=z-a;dt=dz\] for\[t=1,z=1+a\] for\[t=x,z=x+a\] \[\therefore \]\[I=\int_{1+a}^{x+a}{\frac{{{e}^{z-a}}}{z}dz}\] \[={{e}^{-a}}\int_{1+a}^{x+a}{\frac{{{e}^{z}}}{z}dz\equiv {{e}^{-a}}}\int_{1+a}^{x+a}{\frac{{{e}^{t}}}{t}dt}\] \[I={{e}^{-a}}\left[ \int_{1+a}^{1}{\frac{{{e}^{t}}}{t}dz\equiv {{e}^{-a}}}\int_{1+a}^{x+a}{\frac{{{e}^{t}}}{t}dt} \right]\] \[={{e}^{-a}}\left[ \int_{1}^{1+a}{\frac{{{e}^{t}}}{t}dt+}\int_{1}^{x+a}{\frac{{{e}^{t}}}{t}dt} \right]\] \[={{e}^{-a}}[-F(1+a)+F(x+a)]\] (By the definition of F(x)) \[={{e}^{-a}}[F(x+a)-F(1+a)]\]
You need to login to perform this action.
You will be redirected in
3 sec
