given-f-x-f-x-4-x-R-and-5-7-f-x-dx-p-what-is-2-10-f-x-dx-

Question Number 78829 by jagoll last updated on 21/Jan/20

given f (x) = f (x + 4) \forall x \in R

and \underset{5}{\int^{7}} f (x) dx = p . what is

\underset{2}{\int^{10}} f (x) dx ?

Commented by jagoll last updated on 21/Jan/20

i^{'} m solve by mr W

\underset{5}{\int^{7}} f (x) dx = \underset{5}{\int^{7}} f (x - 3) d (x - 3) =

\underset{2}{\int^{4}} f (x) dx = p

now \underset{2}{\int^{10}} f (x) dx = \underset{2}{\int^{4}} f (x) dx + \underset{4}{\int^{6}} f (x) dx + \underset{6}{\int^{8}} f (x) dx

+ \underset{8}{\int^{10}} f (x) dx = 4 p

Commented by jagoll last updated on 21/Jan/20

mr W

my answer correct ?

Commented by mr W last updated on 21/Jan/20

t h e b a s i c i s :

f o r a p e r i o d i c f u n c t i o n f (x) w i t h

p e r i o d T, i . e . f (x) = f (x + T), t h e

i n t e g r a l o v e r a l e n g t h o f T i s a l w a y s

t h e s a m e, i n d e p e d e n t l y f r o m t h e

s t a r t p o i n t o f t h e i n t e g r a l .

s e e d i a g r a m .

e . g . i f t h e p e r i o d i s 4, i . e . f (x) = f (x + 4),

w e h a v e

\int_{1}^{5} f (x) d x = \int_{3}^{7} f (x) d x = \int_{a}^{a + 4} f (x) = \int_{0}^{4} f (x) d x

w e h a v e a l s o

\int_{1}^{13} f (x) d x = \int_{3}^{15} f (x) d x = \int_{a}^{a + 12} f (x) = 3 \int_{0}^{4} f (x) d x

e t c .

k e e p i n m i n d : w e s h o u l d a l w a y t a k e

o n e o r m o r e “ {w h o l e}^{″} p e r i o d s!

e . g . \int_{1}^{4} f (x) d x \neq \int_{3}^{6} f (x) d x

Commented by mr W last updated on 21/Jan/20

Commented by mr W last updated on 21/Jan/20

t h e r e f o r e y o u r a n s w e r i s n o t c o r r e c t .

a n d f o r a f u n c t i o n w i t h p e r i o d 4, i f

o n l y \int_{5}^{7} f (x) d x i s g i v e n, {i t}^{'} s n o t p o s s i b l e

t o f i n d o u t \int_{2}^{10} f (x) d x .

Commented by mr W last updated on 21/Jan/20

\underset{5}{\int^{7}} f (x) dx = \underset{5}{\int^{7}} f (x - 4) d (x - 4) d x = \underset{1}{\int^{3}} f (x) dx = p

\underset{5}{\int^{7}} f (x) dx \neq \underset{5}{\int^{7}} f (x - 3) d (x - 3) d x \neq \underset{2}{\int^{4}} f (x) d x

\underset{2}{\int^{10}} f (x) dx = 2 \underset{1}{\int^{5}} f (x) d x = 2 (\int_{1}^{3} f (x) d x + \int_{3}^{5} f (x) d x)

= 2 (p + \int_{3}^{5} f (x) d x)

= 2 (p + ? ? ?)

Commented by jagoll last updated on 21/Jan/20

oo i understand sir . because f (x) is periodic

function with periode 4, so f (x \pm 4) = f (x)

thanks you sir .

Commented by mr W last updated on 21/Jan/20

{t h a t}^{'} s r i g h t!

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