If the solve the problem
Question:

Let $f(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{g}(\mathrm{t}) \mathrm{dt}$, where $\mathrm{g}$ is a non-zero

even function. If $f(\mathrm{x}+5)=\mathrm{g}(\mathrm{x})$, then $\int^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}$ 

equals-

  1. $\int_{x+5}^{5} g(t) d t$

  2. $5 \int_{x+5}^{5} g(t) d t$

  3. $\int_{5}^{x+5} g(t) d t$

  4. $2 \int_{5}^{x+5} g(t) d t$


Correct Option: 1

Solution:

$f(x)=\int_{0}^{x} g(t) d t$

$f(-\mathrm{x})=\int_{0}^{-\mathrm{x}} \mathrm{g}(\mathrm{t}) \mathrm{dt}$

put $t=-\mathbf{u}$

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

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

$\Rightarrow f(-\mathrm{x})=-f(\mathrm{x})$

$\Rightarrow f(\mathrm{x})$ is an odd function

Also ƒ(5 + x) = g(x)

ƒ(5 – x) = g(–x) = g(x) = ƒ(5 + x)

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

Now

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

$\mathrm{t}=\mathrm{u}+5$

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

$=\int_{-5}^{x-5} g(u) d u$

$=\int_{-5}^{x-5} f^{\prime}(u) d u$

$=f(\mathrm{x}-5)-f(-5)$

$=-f(5-\mathrm{x})+f(5)$

$=f(5)-f(5+\mathrm{x})$

$=\int_{5+x}^{5} f^{\prime}(t) d t=\int_{5+x}^{5} g(t) d t$ 

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