Question

Let $f$ be a twice differentiable function such that $f^{\prime \prime }\left(x\right)+f^{\prime }\left(x\right)=2e^{-x},f\left(0\right)=0$, and $f^{\prime }\left(0\right)=-2.$
Then the area(in sq. units) of region enclosed by $y=f\left(x\right)$ and $x-$axis is

• A. 2.0
• B. 2
• C. 2.00
• D. 02

Answer 1

Answer: (A), (B), (C), (D)

Given: $f^{\prime \prime }\left(x\right)+f^{\prime }\left(x\right)=2e^{-x}$
Integrating w.r.t. $x,$ we get
$f^{\prime }\left(x\right)+f\left(x\right)=-2e^{-x}+C$
Put $x=0,$ we get $C=0$
$\therefore f^{\prime }\left(x\right)+f\left(x\right)=-2e^{-x}$
$\Rightarrow \frac{dy}{dx}+y=-2e^{-x}$
$I.F.=e^x$
Thus, general solution is
$y\cdot e^x=\int -2e^{-x}\cdot e^{x}dx+K$
$\Rightarrow y{e}^x=-2x+K$
Again put $x=0,$ we get $K$=0
$\therefore y=-2x{e}^{-x}$
Now, required area is
$A=\left|{\int }_0^{\infty }-2x{e}^{-x}dx\right|$
$=|-2{[-x{e}^{-x}-e^{-x}]}_0^{\infty }|$
$=2$ square units.