Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If $\int e^x\left(f(x)-f^{\prime}(x)\right)=g(x)+C$, then $\int e^x f^{\prime}(x) d x=$
MathematicsIndefinite IntegrationAP EAMCETAP EAMCET 2022 (06 Jul Shift 1)
Options:
  • A $\frac{1}{2}\left[e^x f(x)-g(x)\right]+C$
  • B $\frac{1}{2}\left[e^x f(x)+g(x)\right]+C$
  • C $\frac{e^x f^{\prime}(x)+g(x)}{2}+C$
  • D $\frac{1}{2}\left[e^x f(x)+e^x g(x)\right]+C$
Solution:
1206 Upvotes Verified Answer
The correct answer is: $\frac{1}{2}\left[e^x f(x)-g(x)\right]+C$
We are given that
$$
\begin{aligned}
& \int e^x\left[f(x)-f^{\prime}(x)\right] d x=g(x)+c \\
& \Rightarrow \int e^x f(x)-\int e^x f^{\prime}(x)=g(x)+c \\
& \Rightarrow \int e^x \cdot f(x) d x=\int e^x f^{\prime}(x)-g(x)+c \\
& \text { Now } \int e^x \cdot f^{\prime}(x) d x= \\
& f(x) \cdot e^x-\int f^{\prime}(x) \cdot e^x d x-g(x)+c \\
& \Rightarrow 2 \int e^x \cdot f^{\prime}(x) d x=e^x \cdot f(x)-g(x)+c \\
& \Rightarrow \int e^x \cdot f^{\prime}(x) d x=\frac{1}{2}\left[e^x \cdot f(x)-g(x)\right]+c
\end{aligned}
$$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.