If $\int e^x\left(f(x)-f^{\prime}(x)\right)=g(x)+C$, then $\int e^x f^{\prime}(x) d x=$

If $\int e^x\left(f(x)-f^{\prime}(x)\right)=g(x)+C$, then $\int e^x f^{\prime}(x) d x=$
  1. $\frac{1}{2}\left[e^x f(x)-g(x)\right]+C$
  2. $\frac{1}{2}\left[e^x f(x)+g(x)\right]+C$
  3. $\frac{e^x f^{\prime}(x)+g(x)}{2}+C$
  4. $\frac{1}{2}\left[e^x f(x)+e^x g(x)\right]+C$

Solution

We are given that $ \begin{aligned} & \int e^x\left[f(x)-f^{\prime}(x)\right] d x=g(x)+c \\ & \Rightarrow \int e^x f(x)-\int e^x f^{\prime}(x)=g(x)+c \\ & \Rightarrow \int e^x \cdot f(x) d x=\int e^x f^{\prime}(x)-g(x)+c \\ & \text { Now } \int e^x \cdot f^{\prime}(x) d x= \\ & f(x) \cdot e^x-\int f^{\prime}(x) \cdot e^x d x-g(x)+c \\ & \Rightarrow 2 \int e^x \cdot f^{\prime}(x) d x=e^x \cdot f(x)-g(x)+c \\ & \Rightarrow \int e^x \cdot f^{\prime}(x) d x=\frac{1}{2}\left[e^x \cdot f(x)-g(x)\right]+c \end{aligned} $

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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