Indefinite Integration 1 Question 79
80. If $f$ and $g$ are continuous functions on $[0, a]$ satisfying $f(x)=f(a-x)$ and $g(x)+g(a-x)=2$, then show that
$$ \int _0^{a} f(x) g(x) d x=\int _0^{a} f(x) d x . $$
Show Answer
Solution:
- Let $I=\int _0^{a} f(x) \cdot g(x) d x$
$$ \begin{array}{cc} & I=\int _0^{a} f(a-x) \cdot g(a-x) d x=\int _0^{a} f(x) \cdot{2-g(x)} d x \\ & \quad[\because f(a-x)=f(x) \text { and } g(x)+g(a-x)=2] \\ & =\quad \int _0^{a} f(x) d x-\int _0^{a} f(x) g(x) d x \\ \Rightarrow & I=2 \int _0^{a} f(x) d x-I \\ \therefore & \quad 2 I=2 \int _0^{a} f(x) d x \\ \therefore & \quad \int _0^{a} f(x) g(x) d x=\int _0^{a} f(x) d x \end{array} $$
