• # question_answer A differentiable function satisfies$f'(x)=f(x)+2{{e}^{x}}$ with initial conditions $f(0)=0$. The area enclosed by f(x) and the x-axis is equal to A)  1                     B)  2 C)  1                                 D)  4

Correct Answer: B

Solution :

Let $f(x)=y$ $\Rightarrow \,\,\frac{dy}{dx}-y=2{{e}^{x}};$ Integrating factor $={{e}^{-x}}$ $\therefore \,\,y.{{e}^{-x}}=\int_{{}}^{{}}{2{{e}^{x}}{{e}^{-x}}\,dx=2x+C}$ Now, $y(0)=0\Rightarrow \,C=0$ $\Rightarrow \,f(x)\,=y=2x{{e}^{x}}$ $\therefore \,\,\frac{dy}{dx}\,=2[x{{e}^{x}}+{{e}^{x}}]\,=0$ $\Rightarrow \,x=-1$ $A=2\int\limits_{-\infty }^{0}{x{{e}^{x}}dx=2\left[ x{{e}^{x}}-\int\limits_{-\infty }^{0}{{{e}^{x}}dx} \right]}$ $=2\left[ x{{e}^{x}}-{{e}^{x}} \right]_{-\infty }^{0}=2\,(-1)\,-(0-0)\,]\,=-2$ Hence area is 2.

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