JEE Main 2014 — Mathematics Calculus

Given $\displaystyle F(x)={\int }_{1}^{x}\frac{{e}^{t}}{t}dt$F(x)=∫1x​tet​dt

Let, $\displaystyle I={\int }_{1}^{x}\frac{{e}^{t}}{t+a}dt$I=∫1x​t+aet​dt

Let, $t+a=y$t+a=y $\Rightarrow dt=dy$⇒dt=dy

Also, $t=1\Rightarrow y=1+a$t=1⇒y=1+a and $t=x\Rightarrow y=x+a$t=x⇒y=x+a

$$\therefore I={\int }_{1+a}^{x+a}\frac{{e}^{y-a}}{y}dy$$∴I=∫1+ax+a​yey−a​dy

$$\Rightarrow I={e}^{-a}{\int }_{1+a}^{x+a}\frac{{e}^{y}}{y}dy$$⇒I=e−a∫1+ax+a​yey​dy

Using $\displaystyle {\int }_{a}^{b}f(x)dx={\int }_{a}^{b}f(t)dt,$∫ab​f(x)dx=∫ab​f(t)dt, we get

$$I={e}^{-a}{\int }_{1+a}^{x+a}\frac{{e}^{t}}{t}dt$$I=e−a∫1+ax+a​tet​dt

$$\Rightarrow I={e}^{-a}[{\int }_{1}^{1+a}\frac{{e}^{t}}{t}dt+{\int }_{1+a}^{x+a}\frac{{e}^{t}}{t}dt-{\int }_{1}^{1+a}\frac{{e}^{t}}{t}dt]$$⇒I=e−a[∫11+a​tet​dt+∫1+ax+a​tet​dt−∫11+a​tet​dt]

Using $\displaystyle {\int }_{a}^{b}f(x)dx+{\int }_{b}^{c}f(x)dx={\int }_{a}^{c}f(x)dx,a<c<b,$∫ab​f(x)dx+∫bc​f(x)dx=∫ac​f(x)dx,a<c<b,

$$\Rightarrow I={e}^{-a}[{\int }_{1}^{x+a}\frac{{e}^{t}}{t}dt-{\int }_{1}^{1+a}\frac{{e}^{t}}{t}dt]$$⇒I=e−a[∫1x+a​tet​dt−∫11+a​tet​dt]

Using, the given relation, we get

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