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Question

Mathematics Question on General and Particular Solutions of a Differential Equation

Let T > 0 be a fixed real number. Suppose, f is a continuous function such that for all xR.f(x+T)=f(x).IfI=0Tf(x)dxx \in R. f(x+T)=f(x). \, If \, I = \int_0^T f(x)dx then the value of 33+3Tf(2x)dx\int_3^{3+3T} f(2x)dx

A

32I\frac{3}{2}I

B

I

C

3I

D

6I

Answer

3I

Explanation

Solution

The correct answer is :3I
Given that;
I=0Tf(x)dxI=\int^T_0f(x)dx, ff is continuous function such that xRx\in{R},f(x+T)=f(x)f(x+T)=f(x), 8(T>0)8(T>0)
To find;33+3Tf(2x)dx\int^{3+3T}_{3}f(2x)dx
take 2x=t  dx=dt22x=t\space \therefore dx=\frac{dt}{2}
12bb(1+T)f(t)dt=620Tf(t)dt\therefore \frac{1}{2}\int^{b(1+T)}_{b}f(t)dt=\frac{6}{2}\int^T_0f(t)dt
=30Tf(x)dt=3\int^T_0f(x)dt
=3I=3I
continuous function