Let $f(x)=\int_{0}^{x} g(t) d t$, where $g$ is a non-zero even function. If
$f(x+5)=g(x)$, then $\int_{0}^{x} f(t) d t$ equals :
Correct Option: 1,
$f(x)=\int_{0}^{x} g(g) d t$............(i)
$\because g$ is a non-zero even function.
$\therefore g(-x)=g(x)$,...........(ii)
Given, $f(x+5)=g(x)$...........(iii)
From (i) $f^{\prime}(x)=g(x)$
Let, $I=\int_{0}^{x} f(t) d t$,
Put $t=\lambda-5 \Rightarrow I=\int_{5}^{x+5} f(\lambda-5) d \lambda$
$\because f(x+5)=\mathrm{g}(x)$
$\Rightarrow f(-x+5)=g(-x)=g(x)$ ...(iv)
$I=\int_{5}^{x+5} f(\lambda-5) d \lambda$
$\because f(0)=0, g(x)$ is even $\Rightarrow f(x)$ is odd
$\therefore I=\int_{5}^{x+5}-f(5-\lambda) d \lambda$
$\Rightarrow I=\int_{5}^{x+5} g(\lambda) d \lambda=\int_{x+5}^{5} g(t) d t$ (from (iv))
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