R-gt-0-0-lt-lt-f-m-sin-t-dt-f-r-sin-2-t-dt-f-m-pi-6-5pi-6-f-m-pi-6-pi-2-f-r-pi-6-5pi-6-f-r-pi-6-pi-

Question Number 1673 by 123456 last updated on 31/Aug/15

ω \in R, ω > 0

0 < α < β

f_{m} (α, β) = \frac{ω}{β - α} \underset{α / ω}{\int^{β / ω}} \sin (ω t) d t

f_{r} (α, β) = \sqrt{\frac{ω}{β - α} \underset{α / ω}{\int^{β / ω}} \sin^{2} (ω t) d t}

f_{m} (\frac{π}{6}, \frac{5 π}{6}) \overset{?}{=} f_{m} (\frac{π}{6}, \frac{π}{2})

f_{r} (\frac{π}{6}, \frac{5 π}{6}) \overset{?}{=} f_{r} (\frac{π}{6}, \frac{π}{2})

Answered by Yozzian last updated on 31/Aug/15

U n d e r t h e i n t e g r a l w i t h i n t h e g i v e n

e x p r e s s i o n s o f f_{m} (α, β) a n d f_{r} (α, β)

ω i s c o n s t a n t . S o w e c a n i n t e g r a t e

a s u s u a l w i t h ω > 0 .

\int_{α / ω}^{β / ω} s i n ω t d t = - \frac{1}{ω} c o s (ω t) ∣_{α / ω}^{β / ω}

r h s = \frac{- 1}{ω} (c o s β - c o s α)

∴ f_{m} (α, β) = \frac{c o s α - c o s β}{β - α}

W h e n α = π / 6 a n d β = 5 π / 6

f_{m} (\frac{π}{6}, \frac{5 π}{6}) = \frac{c o s (\frac{π}{6}) - c o s (\frac{5 π}{6})}{π (\frac{5}{6} - \frac{1}{6})}

= \frac{\frac{\sqrt{3}}{2} - (- \frac{\sqrt{3}}{2})}{\frac{2 π}{3}}

f_{m} (\frac{π}{6}, \frac{5 π}{6}) = \frac{3 \sqrt{3}}{2 π} \dots \dots \dots \dots \dots \dots (1)

W h e n α = \frac{π}{6} a n d β = \frac{π}{2} w e o b t a i n

f_{m} (\frac{π}{6}, \frac{π}{2}) = \frac{c o s (\frac{π}{6}) - c o s (\frac{π}{2})}{π (\frac{1}{2} - \frac{1}{6})}

= \frac{\sqrt{3} / 2 - 0}{π / 3}

f_{m} (\frac{π}{6}, \frac{π}{2}) = \frac{3 \sqrt{3}}{2 π} \dots \dots \dots \dots \dots \dots . . (2)

U p o n c o m p a r i s o n o f r e s u l t s (1)

a n d (2), w e s e e t h a t t h e y a r e e q u a l

i n v a l u e .

H e n c e, f_{m} (\frac{π}{6}, \frac{5 π}{6}) = f_{m} (\frac{π}{6}, \frac{π}{2}) .

T h e n e x t i n t e g r a l a l s o h a s ω

c o n s t a n t w i t h r e s p e c t t o t h e

v a r i a b l e u n d e r i n t e g r a t i o n t .

∴ \int_{α / ω}^{β / ω} {s i n}^{2} ω t d t = \frac{1}{2} \int_{α / w}^{β / ω} (1 - c o s (2 ω t)) d t

r h s = \frac{1}{2} (t - \frac{1}{2 ω} s i n (2 ω t)) ∣_{α / ω}^{β / ω}

r h s = \frac{1}{2} (\frac{β - α}{ω} + \frac{1}{2 ω} (s i n (2 α) - s i n (2 β)))

r h s = \frac{2 (β - α) + s i n 2 α - s i n 2 β}{4 ω}

∴ f_{r} (α, β) = \sqrt{\frac{2 (β - α) + s i n 2 α - s i n 2 β}{4 (β - α)}}

W h e n α = \frac{π}{6} a n d β = \frac{5 π}{6} w e g e t

f_{r} (\frac{π}{6}, \frac{5 π}{6}) = \sqrt{\frac{1}{2} + \frac{s i n \frac{π}{3} - s i n \frac{5 π}{3}}{\frac{2 π}{3}}}

= \sqrt{\frac{1}{2} + \frac{\frac{\sqrt{3}}{2} - (- \frac{\sqrt{3}}{2})}{2 π / 3}}

= \sqrt{\frac{1}{2} + \frac{3 \sqrt{3}}{2 π}}

f_{r} (\frac{π}{6}, \frac{5 π}{6}) = \sqrt{\frac{π + 3 \sqrt{3}}{2 π}} . \dots \dots . (3)

W h e n α = \frac{π}{6} a n d β = \frac{π}{2} w e h a v e

f_{r} (\frac{π}{6}, \frac{π}{2}) = \sqrt{\frac{1}{2} + \frac{s i n \frac{π}{3} - s i n π}{π / 3}}

= \sqrt{\frac{π + 3 \sqrt{3}}{2 π}} . \dots \dots . . (4)

O n c o m p a r i n g t h e r e s u l t s (3) a n d

(4) w e s e e t h a t t h e y a r e e q u a l i n

v a l u e . H e n c e,

f_{r} (\frac{π}{6}, \frac{5 π}{6}) = f_{r} (\frac{π}{6}, \frac{π}{2}) .

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