The integration of a periodic function involves some unique properties and approaches due to the function’s repeating nature. A function $f(x)$ is said to be periodic with period $T$ if for all $x$ in the domain of $f$, $f(x + T) = f(x)$. This periodicity can significantly simplify the integration process over intervals that are multiples of the period $T$.<\/p>\n\n\n\n
(1) Integral over one period<\/strong>: The integral of a periodic function over one period is a constant value that does not depend on the starting point of the interval. Mathematically, this is expressed as: Proof<\/strong><\/p>\n\n\n\n $\\int_{0}^{T} f(x)\\,dx = \\int_{0}^{a} f(x)\\,dx + \\int_{a}^{T} f(x)\\,dx$ (2) Integral over multiple periods<\/strong>: If you integrate a periodic function over an interval that spans multiple periods, the result is simply the integral over one period multiplied by the number of periods covered. For an interval $[a, b]$ that spans $n$ periods: We can derive below formula based on above<\/p>\n\n\n\n $$ $$ $$ Example 1<\/strong><\/p>\n\n\n\n $$ Solution<\/strong> So Question 1<\/strong> Solution<\/strong> Hence this is independent of a<\/p>\n\n\n\n Question 2<\/strong><\/p>\n\n\n\n Assertion :If $I=\\int_{0}^{T} f(x)\\,dx $, then $\\int_{3}^{3+3T} f(2x)\\,dx =3I$ (a)Both Assertion and Reason are correct and Reason is the correct explanation for Assertion Solution<\/strong> Other Integration Related Articles<\/p>\n\n\n\n The integration of a periodic function involves some unique properties and approaches due to the function’s repeating nature. A function $f(x)$ is said to be periodic with period $T$ if for all $x$ in the domain of $f$, $f(x + T) = f(x)$. This periodicity can significantly simplify the integration process over intervals that are […]<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"set","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[24,498],"tags":[],"class_list":["post-9022","post","type-post","status-publish","format-standard","hentry","category-general","category-maths"],"yoast_head":"\n
$$
\\int_{a}^{a+T} f(x)\\,dx = \\int_{0}^{T} f(x)\\,dx =constant
$$
where $T$ is the period of the function and $a$ is any real number.<\/p>\n\n\n\n
$= \\int_{0}^{a} f(x)\\,dx + F(T) – F(a)$
$=\\int_{0}^{a} f(x)\\,dx + F(a+T) – F(a) – (F(a+T) – F(p) ) $
$=\\int_{0}^{a} f(x)\\,dx + \\int_{a}^{a+T} f(x)\\,dx – \\int_{T}^{a+T} f(x)\\,dx $
Now Calulate $\\int_{T}^{a+T} f(x)\\,dx $
Now let x = p + T
then dx= dp
$\\int_{T}^{a+T} f(x)\\,dx = \\int_{0}^{a} f((p+T))\\,dt $
Since this is periodic with period T , f(p+T) = f(p)
Therefore
$=\\int_{0}^{a} f(p)\\,dp $
$=\\int_{0}^{a} f(x)\\,dx $
Subsituting back, we have
$\\int_{0}^{T} f(x)\\,dx = \\int_{0}^{a} f(x)\\,dx + \\int_{a}^{a+T} f(x)\\,dx – \\int_{0}^{a} f(x)\\,dx=\\int_{a}^{a+T} f(x)\\,dx $<\/p>\n\n\n\n
$$
\\int_{a}^{a+nT} f(x)\\,dx = n \\cdot \\int_{a}^{a+T} f(x)\\,dx,
$$
where $n = \\frac{b-a}{T}$ (assuming $b-a$ is an exact multiple of $T$).<\/p>\n\n\n\n
\\int_{0}^{nT} f(x)\\,dx = n \\cdot \\int_{0}^{T} f(x)\\,dx,
$$<\/p>\n\n\n\n
\\int_{nT}^{mT} f(x)\\,dx = (m-n) \\cdot \\int_{0}^{T} f(x)\\,dx,
$$<\/p>\n\n\n\n
\\int_{a+nT}^{b+nT} f(x)\\,dx = \\int_{a}^{b} f(x)\\,dx,
$$<\/p>\n\n\n\n
\\int_{0}^{6\\pi} \\sin(x)\\,dx
$$<\/p>\n\n\n\n
The function $f(x) = \\sin(x)$, is a periodic with period $T = 2\\pi$.<\/p>\n\n\n\n
$$
\\int_{0}^{6\\pi} \\sin(x)\\,dx= 3 \\times \\int_{0}^{2\\pi} \\sin(x)\\,dx= 3 \\times 0 =0
$$<\/p>\n\n\n\nSolved Questions<\/h2>\n\n\n\n
h(x) is continuous periodic function with period T, then the integral
$$
I= \\int_{a}^{a+T} h(x)\\,dx
$$
is ?
(a) equal to 2a
(b) Independent of a
(c) 4a
(d) None of these<\/p>\n\n\n\n
we know by property
$$
\\int_{a}^{a+T} f(x)\\,dx = \\int_{0}^{T} f(x)\\,dx =constant
$$<\/p>\n\n\n\n
Reason:f(x) is periodic function with time period T<\/p>\n\n\n\n
(b) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
(c)Assertion is correct but Reason is incorrect
(d)Both Assertion and Reason are incorrect<\/p>\n\n\n\n
$\\int_{3}^{3+3T} f(2x)\\,dx $
let 2x=p, then $dx = \\frac {dp}{2} $
$\\int_{3}^{3+3T} f(2x)\\,dx = \\frac {1}{2} \\int_{6}^{6+6T} f(p)\\,dp$
$=\\frac {1}{2} \\int_{6}^{6+6T} f(x)\\,dx=3 \\int_{0}^{T} f(x) dx=3I $
(a) is the correct answer<\/p>\n\n\n\nIntegration of modulus function<\/a><\/td> Integration of tan x<\/a><\/td> <\/td> <\/td><\/tr> Integration of sec x<\/a><\/td> Integration of greatest integer function<\/a><\/td> <\/td> <\/td><\/tr> Integration of trigonometric functions<\/a><\/td> Integration of cot x<\/a><\/td> <\/td> <\/td><\/tr> Integration of cosec x<\/a><\/td> Integration of sinx<\/a><\/td> <\/td> <\/td><\/tr> Integration of irrational functions<\/a><\/td> integration of root x<\/a><\/td> <\/td> <\/td><\/tr> Integration of fractional part of x<\/a><\/td> Integration of cos square x<\/a><\/td> <\/td> <\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"