Let f be a periodic continuous function with period T$$0$$. If $$I=$$∫$$0T$$ $$f(x)dx$$ Then the value of $$I1=$$∫$$44+4T$$ $$f(3x)dx$$ is

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Let f be a periodic continuous function with period T>$$0$$. If $$I=$$∫$$0T$$ $$f(x)dx$$ Then the value of $$I1=$$∫$$44+4T$$ $$f(3x)dx$$ is

Infinity Learn · Accepted Answer

Put $$3x=y$$ in $$I1$$ $$I1=13$$∫$$1212+12T$$ $$f(y)dy=13$$∫$$12T$$ $$f(y)dy+$$∑$$K=111$$ ∫$$KT(K+1)T$$ $$f(y)dy+$$∫$$12T12+12T$$ $$f(y)dyBut$$ ∫$$KT(K+1)T$$ $$f(y)dy=$$∫$$0T$$ $$f(u)du$$ , (Put$$ $$KT+u=y$$ or use Property $$17) and ∫$$12T12+12T$$ $$f(y)dy=$$∫$$012$$ $$f(u)duHence$$ $$I1=13$$∫$$12T$$ $$f(y)dy+11I+$$∫$$012$$ $$f(u)du$$               $$=13$$∫$$0T$$ $$f(y)dy+11I=13$$×$$12I=4I$$.

Let $f$ be a periodic continuous function with period
$T > 0 .$ If $I = \int_{0}^{T} f (x) d x$ Then the value of $I_{1} = \int_{4}^{4 + 4 T} f (3 x) d x$ is

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Let f be a periodic continuous function with period T>0. If I=∫0T f(x)dx Then the value of I1=∫44+4T f(3x)dx is

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Let $f$ be a periodic continuous function with period
$T > 0 .$ If $I = \int_{0}^{T} f (x) d x$ Then the value of $I_{1} = \int_{4}^{4 + 4 T} f (3 x) d x$ is