Question

Let $\int e^x\left\{f\right(x)-f^{\prime }(x\left)\right\}dx=\varphi \left(x\right)$. Then $\int e^{x}f\left(x\right)dx$ is
(where $c$ is constant of integration)

• A. $\varphi \left(x\right)+e^{x}f\left(x\right)+c$
• B. $\varphi \left(x\right)-e^{x}f\left(x\right)+c$
• C. $\frac{1}{2}\left\{\varphi \right(x)+e^{x}f(x\left)\right\}+c$
• D. $\frac{1}{2}\left\{\varphi \right(x)+e^x{f}^{\prime }(x\left)\right\}+c$

Answer 1

The correct option is C $\frac{1}{2}\left\{\varphi \right(x)+e^{x}f(x\left)\right\}+c$

Here, $\int e^x\left\{f\right(x)-f^{\prime }(x\left)\right\}dx=\varphi \left(x\right)\cdots \left(i\right)$ and $\int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+c^{\prime }\cdots \left(ii\right)$
On adding $\left(i\right)$ and $\left(ii\right)$, we get $2\int e^{x}f\left(x\right)dx=\varphi \left(x\right)+e^{x}f\left(x\right)+c^{\prime }$
$\Rightarrow \int e^{x}f\left(x\right)dx=\frac{1}{2}\left\{\varphi \right(x)+e^{x}f(x\left)\right\}+c$