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Question Number 173350 by Muktarr last updated on 10/Jul/22

	Answered by aleks041103 last updated on 10/Jul/22
	$r.m.s. = root mean square RMS(f(x))=(√(⟨f^( 2) (x)⟩))=(√((1/T)∫_x_0 ^(x_0 +T) f^( 2) (x) dx)) where f(x)=f(x+kT), k∈Z in this case, T=1 and f(x)=2+4frac(x), where frac(x):=x−⌊x⌋ ⇒f_(rms) =(√((1/1)∫_0 ^( 1) (2+4frac(x))^2 dx)) ∫_0 ^( 1) (2+4frac(x))^2 dx=4∫_0 ^1 (1+2x)^2 dx= =2∫_0 ^1 (1+2x)^2 d(1+2x)= =(2/3)[(1+2x)^3 ]_0 ^1 =(2/3)((1+2.1)^3 −(1+2.0)^3 )= =(2/3)(27−1)=((52)/3) ⇒f_(rms) =(√((52)/3))$
	$r . m . s . = r o o t m e a n s q u a r e$ $R M S (f (x)) = \sqrt{⟨ f^{2} (x) ⟩} = \sqrt{\frac{1}{T} \underset{x_{0}}{\int^{x_{0} + T}} f^{2} (x) d x}$ $w h e r e f (x) = f (x + k T), k \in Z$ $i n t h i s c a s e, T = 1$ $a n d f (x) = 2 + 4 f r a c (x), w h e r e f r a c (x) := x - ⌊ x ⌋$ $\Rightarrow f_{r m s} = \sqrt{\frac{1}{1} \int_{0}^{1} {(2 + 4 f r a c (x))}^{2} d x}$ $\int_{0}^{1} {(2 + 4 f r a c (x))}^{2} d x = 4 \int_{0}^{1} {(1 + 2 x)}^{2} d x =$ $= 2 \int_{0}^{1} {(1 + 2 x)}^{2} d (1 + 2 x) =$ $= \frac{2}{3} {[{(1 + 2 x)}^{3}]}_{0}^{1} = \frac{2}{3} ({(1 + 2.1)}^{3} - {(1 + 2.0)}^{3}) =$ $= \frac{2}{3} (27 - 1) = \frac{52}{3}$ $\Rightarrow f_{r m s} = \sqrt{\frac{52}{3}}$
	Commented by Tawa11 last updated on 11/Jul/22

	$Great sir$
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