Indefinite Integration 1 Question 8
9. Let $f$ and $g$ be continuous functions on $[0, a]$ such that $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$, then $\int _0^{a} f(x) g(x) d x$ is equal to
(2019 Main, 12 Jan I)
(a) $4 \int _0^{a} f(x) d x$
(b) $\int _0^{a} f(x) d x$
(c) $2 \int _0^{a} f(x) d x$
(d) $-3 \int _0^{a} f(x) d x$
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Solution:
- Let $I=\int _0^{a} f(x) g(x) d x$
$$ =\int _0^{a} f(a-x) g(a-x) d x $$
$$ \because \int _0^{a} f(x) d x=\int _0^{a} f(a-x) d x $$
$\Rightarrow \quad I=\int _0^{a} f(x)[4-g(x)] d x$
$$ [\because f(x)=f(a-x) \text { and } g(x)+g(a-x)=4] $$
$$ =\int _0^{a} 4 f(x) d x-\int _0^{a} f(x) g(x) d x $$
$\Rightarrow \quad I=4 \int _0^{a} f(x) d x-I$
[from Eq. (i)]
$\Rightarrow \quad 2 I=4 \int _0^{a} f(x) d x \Rightarrow \quad I=2 \int _0^{a} f(x) d x$.
