Question

If $y=f_1\left(x\right)$ and $y=f_2\left(x\right)$ are two solutions of the equation $ydx+dy=-e^x{y}^{2}dy$, where $f_1\left(0\right)=1$ and $f_2\left(0\right)=-1$.

If area enclosed by $y=f_1\left(x\right),y=e^x$ and the x-axis is $A_1$ and $y=f_2\left(x\right),y=-e^x$ and the x-axis is $A_2$ then $A_1+A_2$ is equal to

• A. $6\ln 2$
• B. 3
• C. 6
• D. $3\ln 2$

Answer 1

The correct option is C 6

Area enclosed by $y=f_1\left(x\right),y=e^x$ and the x-axis:

$A_1={\int }_{-\infty }^0{e}^{x}dx+{\int }_0^{\infty }{e}^{\frac{-x}{2}}dx$
$=(e^x{)}_{-\infty }^0+[-2{\left(e^{\frac{-x}{2}}\right)}_0^{\infty }]$
$=1-0-2(0-1)=3$
Area enclosed by $y=f_2\left(x\right);y=-e^x$ and the x-axis.

$A_2={\int }_{-\infty }^0(-e^x)dx+{\int }_0^{\infty }(-e^{\frac{-x}{2}})dx$
$\Rightarrow A_2=3$
$\therefore A_1+A_2=6$