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:

  1. 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)]

$\Rightarrow \quad 2 I=4 \int _0^{a} f(x) d x \Rightarrow \quad I=2 \int _0^{a} f(x) d x$.



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