Question

If \[ \int e^x\{f(x)-f'(x)\}=g(x)+C, \] then \[ \int e^x f'(x)\,dx= \]

• A. \(\dfrac{1}{2}\left[e^x f(x)-g(x)\right]+C\)
• B. \(\dfrac{1}{2}\left[e^x f(x)+g(x)\right]+C\)
• C. \(\dfrac{e^x f'(x)+g(x)}{2}+C\)
• D. \(\dfrac{1}{2}\left[e^x f(x)+e^x g(x)\right]+C\)

Hint:

Use the product derivative \[ \frac{d}{dx}\left(e^x f(x)\right)=e^x f(x)+e^x f'(x) \] to connect integrals involving \(e^x f(x)\) and \(e^x f'(x)\).

Answer 1

The Correct Option is A

Step 1: Use the given integral relation.
Given: \[ \int e^x\{f(x)-f'(x)\}\,dx=g(x)+C \] This means \[ \int e^x f(x)\,dx-\int e^x f'(x)\,dx=g(x)+C \] Let \[ I=\int e^x f'(x)\,dx \] Also let \[ J=\int e^x f(x)\,dx \] Then, \[ J-I=g(x) \] So, \[ J=I+g(x) \]

Step 2: Use derivative of \(e^x f(x)\).
We know that \[ \frac{d}{dx}\left(e^x f(x)\right)=e^x f(x)+e^x f'(x) \] Integrating both sides: \[ e^x f(x)=\int e^x f(x)\,dx+\int e^x f'(x)\,dx \] Therefore, \[ e^x f(x)=J+I \]

Step 3: Substitute \(J=I+g(x)\).
\[ e^x f(x)=I+g(x)+I \] \[ e^x f(x)=2I+g(x) \] Thus, \[ 2I=e^x f(x)-g(x) \] \[ I=\frac{1}{2}\left[e^x f(x)-g(x)\right] \]

Step 4: Final conclusion.
Hence, \[ \int e^x f'(x)\,dx = \boxed{\frac{1}{2}\left[e^x f(x)-g(x)\right]+C} \]