Let f(x) = ∫_0^x g(t) dt, where g is a non-zero even function. If f(x+5)=g(x), then ∫_0^x f(t)dt equals

Correct option is A. ∫x+55g(t)dt
f(x)=∫0^x g(t)dt.. (1)
f(−x)=∫−x^0 g(t)dt
put t=−u dt=−du
f(−x)=∫0^x g(−u)du
f(−x)=−∫0^x g(−u)du
f(x)=−∫0^x g(u)du
f(x)=−f(x)
Here f(x) is odd function
given f(5+x)=g(x)
f(5−x)=g(−x)
f(5−x)=g(x) ( because g(x) is even function)
f(5−x)=f(5+x)
f(5−x)=f(5+x)=g(x)
Now I=∫0^x f(t)dt
put t=u+5 dt=du
I=∫−5^x−5 f(u+5)du=∫−5^x−5 g(u)du=∫−5^x−5 f'(u)du=[f(u)]−5^x−5=f(x−5)−f(−5)
=f(−(5−x))−f(−5)(∵f(5+x)=g(x) f(5+u)=g(u))
f(x)=∫_0^x g(t)dt
differentiate f'(x)=g(x)
f'(u)=g(u)
f(x) is odd function
f(−x)=−f(x)
=f(−(5−x))−f(−5)=−f(5−x)+f(5)=−f(5)−f(5+x)
I=f(5)−f(5+x)=∫_5+x^5 f'(x)d=∫_5+x^5 g(t)dt
f(x)=∫_0^x g(t)dt
f'(x)=g(t)=∫_5+x^5 g(t)dt
Ans.