If g(x) = _0^x 4t,dt, then g(x+) equals:

Question

If $g(x) = \displaystyle\int_0^x \cos4t\,dt$, then $g(x+\pi)$ equals:

Accepted Answer

$$\left( {b,c} \right)g\left( {x + \pi } \right) = \int\limits_0^{x + \pi } {\cos \,4t\,dt} $$

$$ = \int\limits_0^\pi  {\cos 4tdt + \int\limits_\pi ^{\pi  + x} {\cos 4t\,dt} } $$

$$ = g\left( \pi  \right) + \int\limits_0^x {\cos \,4t\,dt} $$

Putting $$t = \pi  + y$$ in second integral, we get

$$\int\limits_x^{\pi  + x} {\cos \,4t\,dt = \int\limits_0^\pi  {\cos \,4t\,dt} } $$

$$ = g\left( \pi  \right) + g\left( x \right) = g\left( x \right) - g\left( \pi  \right)$$

$$\therefore$$ $$\left. {g\left( \pi  \right) = 0} \right)$$

Solutions · 1