State which of the following statement(s) is / are true?

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If m and M are the least and the greatest value of $f (x) i n [a, b], t h e n$ $m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a)$

If $f^{2} (x)$ and $g^{2} (x)$ are integrable on [a, b], then $| \int_{a}^{b} f (x) g (x) d x | \leq {(\int_{a}^{b} f^{2} (x) d x)}^{1 / 2} {(\int_{a}^{b} g^{2} (x) d x)}^{1 / 2}$

If $f (x)$ is continuous on [a,b], then there exists a point $c \in (a, b),$ such that $\int_{a}^{b} f (x) d x = f (c) (b - a)$

If $f (x)$ is periodic function with period T and $a \in R^{+}$ , then $\int_{a}^{a + T} f (x) d x = \int_{0}^{T} f (x) d x$

answer is A, B, C, D.

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Detailed Solution

Sol. : for A
$\begin{array}{l} ∵ m \leq f (x) \leq M \\ \Rightarrow \int_{a}^{b} m d x \leq \int_{a}^{b} f (x) d x \leq \int_{a}^{b} M d x \\ \Rightarrow m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a) \\ \Rightarrow (A) i s c o r r e c t \end{array}$
Sol. : for B
Schwartz ¬– Bunyakovsky inequality
$\begin{array}{l} ∵ m \leq f (x) \leq M \\ \Rightarrow (B) i s c o r r e c t \end{array}$
Sol. : for C
Schwartz ¬– Bunyakovsky inequality
$\begin{array}{l} ∵ m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a) \exists c \in (a, b) s u c h t h a t \\ m < f (c) < M a n d \int_{a}^{b} f (x) d x = (b - a) . f (c) \\ \Rightarrow (C) i s c o r r e c t \end{array}$
Sol. : for D
$\begin{array}{l} I = \int_{a}^{a} f (x) d x \\ P u t x = a + t; d x = d t \\ I = \int_{a}^{T} f (a + t) d x = \int_{a}^{T} f (x) d x a s f (x) i s p e r i o d i c w i t h p e r i o d T \\ \Rightarrow (D) i s c o r r e c t \end{array}$