Definite Integration Question 78
Question 78
- 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
$$ \int_{0}^{a} f(x) g(x) d x=\int_{0}^{a} f(x) d x . $$
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Solution:
- Let $I=\int_{0}^{a} f(x) \cdot g(x) d x$
$$ \begin{array}{cc} & I=\int_{0}^{a} f(a-x) \cdot g(a-x) d x=\int_{0}^{a} f(x) \cdot{2-g(x)} d x \ & \quad[\because f(a-x)=f(x) \text { and } g(x)+g(a-x)=2] \ & =\quad \int_{0}^{a} f(x) d x-\int_{0}^{a} f(x) g(x) d x \ \Rightarrow & I=2 \int_{0}^{a} f(x) d x-I \ \therefore & \quad 2 I=2 \int_{0}^{a} f(x) d x \ \therefore & \quad \int_{0}^{a} f(x) g(x) d x=\int_{0}^{a} f(x) d x \end{array} $$
