Let `f(x)=int_(0)^(x)f ` (t) dt equals

Options

A. `5int_(x+5)^(5)g(t)dt`
B. `int_(5)^(x+5)f(t)dt`
C. `2int_(5)^(x+5)f(t)dt`
D. `int_(x+5)^(5)g(t)dt`

Answer

D

Explanation

Given , f (x) `=int_(0)^(x)f(t)dt`
On replacing x by (-x) , we get
`f(x)=int_(0)^(-x)g (t)dt`
Now , put t =- u , so
`int(-x)=-int_(0)^(x)g(-u)du=-int_(0)^(x)g(u)=-f(x)`
`[:' ` g is an even function]
`rArrf(-x)=-f(xrArr f` an odd function.
Now , it is given that f (x +5) = g (x)
`:.f(5-x)=g(-x)=g(x)=f(x+5)`
[ `:.` g is an even function]
`rArrf(5-x)=f(x+5)` . . . (i)
Let `I=int_(0)^(x)f(t)dt`
Put t = u +5 `rArrt -5=u rArrdt = du`
`:.u=-trArrdu=-dt ` , we get
`I=-int_(5)^(5-x)g(-t)dt=int_(5-x)^(5)f (t) dt`
`[:'-int_(a)^(b)f(x)dx=int_(b)^(a)f (x)` dx and g is an even function]
`I=int_(5-x)^(5)f'(t)dt`
[ by Leibnitz rule f ' (x) = g (x) ]
`=f(5)-f(5-x)=f(5)-f(5+x)` [ from Eq . (i)]
`=int_(5+x)^(5) f ' (t) dt = int_(5+x)^(5)g (t) dt`