Let f(x) = _0^x g(t),dt, where g is a non-zero even function.

Question

Let $f(x) = \displaystyle\int_0^x g(t)\,dt$, where $g$ is a non-zero even function. If $f(x+5) = g(x)$, then $\displaystyle\int_0^x f(t)\,dt$ equals:

Accepted Answer

$$f(x) = \int\limits_0^x {g(t)dt} $$

$$f\left( { - x} \right) = \int\limits_0^{ - x} {g\left( t \right)} dt$$

Put t = -v

= $$ - \int\limits_0^x {g\left( { - v} \right)} dv$$

= $$ - \int\limits_0^x {g\left( v \right)} d(v)$$ [ as g(v) is an even function.]

= - f(x)

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

$$ \therefore $$  f(x) is an odd function.

Given f(x + 5) = g(x)

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

$$ \Rightarrow $$ g(x) = - f(x - 5) [as g(x) is even and f(x) is an odd function]

Replacing x by x + 5, we get

f(x) = - g(x + 5) ......(1)

Now

$$ \int\limits_0^x {f(t)dt} $$

= $$ - \int\limits_0^x {g\left( {t + 5} \right)} dt$$

= $$ - \int\limits_5^{x + 5} {g\left( t \right)} dt$$

= $$\int\limits_{x + 5}^5 {g(t)dt} $$

Solutions · 1