TS EAMCET · Maths · Definite Integration
If \(\int_{-a}^a f(x) d x=\int_0^a f(x) d x+\int_0^a g(x) d x\) then \(g(x)=\)
- A \(-f(x)\)
- B \(f(x)\)
- C \(f(-x)\)
- D \(f(x)+f(-x\)
Answer & Solution
Correct Answer
(C) \(f(-x)\)
Step-by-step Solution
Detailed explanation
Since, \(\int_{-a}^a f(x) d x=2 \int_0^a f(x) d x\) \[ \begin{aligned} & =\int_0^a f(x) d x=f \int_0^a f(x) d x \\ & =\int_0^a f(x) d x=f \int_0^a f(-x) d x \end{aligned} \] Here \(f(-x)=f(x)\) is an even function \[ \Rightarrow \mathrm{g}(\mathrm{x})=\mathrm{f}(-\mathrm{x}) \]
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