Let f\left ( x \right )=\int_{0}^{x}g(t)dt, where g is a non-zero even function. If f(x+5)=g(x), then \int_{0}^{x}f(t)dt equals : 


 

  • Option 1)

    \int_{x+5}^{5}g(t)dt

  • Option 2)

    \int_{5}^{x+5}g(t)dt

  • Option 3)

    2\int_{5}^{x+5}g(t)dt

     

  • Option 4)

    5\int_{x+5}^{5}g(t)dt

Answers (1)


f\left ( x \right )=\int_{0}^{x}g(t)dt

f\left ( -x \right )=\int_{0}^{-x}g(t)dt

put t=-u

         =-\int_{0}^{x}g(-u)du=-\int_{0}^{x}g(u)d(u)=-f(x)

\because g is even function.

\Rightarrow f\left ( -x \right )=-f(x)

\Rightarrow f(x) is an odd function 

Also given

 \\f(5+x)=g(x)\\f(5-x)=g(-x)=g(x)=f(5+x)

\Rightarrow f(5-x)=f(5+x)

Now, let

I=\int_{0}^{x}f(x)dt

t=u+5

I=\int_{-5}^{x-5}f(u+5)du=\int_{-5}^{x-5}g(u)du

   = \int_{-5}^{x-5}{f}'(u)du\: \: \: \: \: \: \: \: from\cdots (1)

  =f(x-5)-f(-5)=-f(5-x)+f(5)

=f(5)-f(5+x)

=\int_{5+x}^{5}{f}'(t)dt=\int_{5+x}^{5}g(t)dt


Option 1)

\int_{x+5}^{5}g(t)dt

Option 2)

\int_{5}^{x+5}g(t)dt

Option 3)

2\int_{5}^{x+5}g(t)dt

 

Option 4)

5\int_{x+5}^{5}g(t)dt

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