SATHEE: Indefinite Integration 1 Question 79

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

$\int_{0}^{a} f (x) g (x) d x = \int_{0}^{a} f (x) d x .$

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Solution:

Let $I = \int_{0}^{a} f (x) \cdot g (x) d x$

$\begin{array}{cc} I = \int_{0}^{a} f (a - x) \cdot g (a - x) d x = \int_{0}^{a} f (x) \cdot 2 - g (x) d x \\ [∵ f (a - x) = f (x) and g (x) + g (a - x) = 2] \\ = \int_{0}^{a} f (x) d x - \int_{0}^{a} f (x) g (x) d x \\ \Rightarrow & I = 2 \int_{0}^{a} f (x) d x - I \\ ∴ & 2 I = 2 \int_{0}^{a} f (x) d x \\ ∴ & \int_{0}^{a} f (x) g (x) d x = \int_{0}^{a} f (x) d x \end{array}$

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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

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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