Question

If $\int e^x\left(f\right(x)-f^{\prime }(x\left)\right)dx=\varphi \left(x\right)$, then $\int e^{x}f\left(x\right)dx=$

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

Answer 1

Answer: (C)

$\frac{1}{2}\left\{\varphi \right(x)+e^{x}f(x\left)\right\}$

$\varphi \left(x\right)=\int e^x\left[f\right(x)-f^{\prime }(x\left)\right]dx$

$\varphi \left(x\right)$ $=\int e^{x}f\left(x\right)dx-\int e^x{f}^{\prime }\left(x\right)dx$

and

$\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^x.f\left(x\right)$

$\because \frac{d}{dx}(e^x.f(x\left)\right)=e^x\left[f\right(x)+f^{\prime }(x\left)\right]$

So, $\int e^{x}f\left(x\right)dx-\int e^x.f^{\prime }\left(x\right)dx=\varphi \left(x\right)$ --(1), (Given equation).

$\int e^{x}f\left(x\right)dx+\int e^x{f}^{\prime }\left(x\right)dx=e^{x}f\left(x\right)$ --(2)

$Eqn\left(1\right)+Eqn\left(2\right)$ now gives us,

$2\int e^{x}f\left(x\right)dx=\theta \left(x\right)+e^{x}f\left(x\right)$

$\int e^{x}f\left(x\right)dx=\frac{1}{2}\left(\varphi \right(x)+e^{x}f(x\left)\right)$