Let x be a continuous variable defined over the interval $$\left( { – \infty ,\,\infty } \right)$$ , and f(x) = e-x-e-x . The integral $${\text{g}}\left( {\text{x}} \right) = \int {{\text{f}}\left( {\text{x}} \right){\text{dx}}} $$ is equal to A. ee-x B. e-e-x C. e-ex D. e-x

The correct answer is $\boxed{\text{D}}$.

The integral of $e^{-x} – e^{-x}$ is $e^{-x}$. This is because the derivative of $e^{-x}$ is $-e^{-x}$, so integrating $-e^{-x}$ gives $e^{-x}$ plus an arbitrary constant. Since we are integrating over the entire real line, the constant of integration must be zero.

Here is a more detailed explanation:

The integral of a function $f(x)$ is the area under the curve $y=f(x)$ between the $x$-axis and the line $x=a$, where $a$ is the upper limit of integration. In this case, $f(x)=e^{-x}-e^{-x}$ and $a=\infty$.

The area under the curve $y=e^{-x}-e^{-x}$ is shown in the following graph:

[asy]
unitsize(1 cm);

draw((0,0)–(10,0));
draw((0,0)–(0,1));

real ticklen=1;
real tickspace=1;
real axisarrowsize=0.14inch;
real tickdown=-0.12inch;
real tickdownlength=-0.12inch;
real wholetickdown=-0.24inch;
real wholetickdownlength=-0.24inch;
real tickdownlengthshort=-0.06inch;
real wholetickdownlengthshort=-0.12inch;
real tickdownbase=0.12inch;
real wholetickdownbase=0.24inch;
real wholetickdownbaseshort=0.36inch;
real t=0;
real dt=1;
real xleft=-2;
real xright=12;
real ybottom=-1;
real ytop=1.2;

label(“$x$”,(xright,0),E);
label(“$y$”,(0,ytop),N);

real i;
for (i=xleft; i<xright; i+=dt) {
draw((i,0)–(i,0.1));
}

for (i=ybottom; i<ytop; i+=0.2) {
draw((0,i)–(0.1,i));
}

draw((0,0)–(xright,0));
draw((0,0)–(0,ytop));

real f(real x) {
return exp(-x)-exp(-x);
}

real g(real x) {
return exp(-x);
}

draw(graph(f,xleft,xright),red);
draw(graph(g,xleft,xright),blue);

draw((xleft,f(xleft))–(xleft,0));
draw((xright,f(xright))–(xright,0));
draw((0,g(0))–(0,ytop));

label(“$e^{-x}$”,(xright,f(xright)),E);
label(“$e^{-x}-e^{-x}$”,(xright,0),E);
label(“$e^{-x}$”,(0,g(0)),N);
[/asy]

The area under the curve is equal to the integral of $f(x)$ from $x=0$ to $x=\infty$. This integral is equal to $e^{-x}$.