First slide

Evaluation of definite integrals

Question

If $g (x) = \int_{0}^{x} \cos 4 t d t$ ,, then $g (x + π)$ equals

Moderate

A

$g (x) \pm g (π)$
B

$g (π) - g (x)$
C

$- g (x)$
D

$g (x) g (π)$

Solution

$\begin{array}{l} g (x + π) = \int_{0}^{x + π} \cos 4 t d t \\ = \int_{0}^{π} \cos 4 t d t + \int_{π}^{x + π} \cos 4 t d t \\ = g (π) + I \end{array}$
where $I = \int_{π}^{x + π} \cos 4 t d t,$ Put $t = π + θ$ so that
$I = \int_{0}^{x} \cos (4 π + 4 θ) d θ = \int_{0}^{x} \cos 4 θ d θ = g (x)$
So $g (x + π) = g (π) + g (x)$ but
$\begin{array}{l} g (π) = \int_{0}^{π} \cos 4 t d t = \frac{1}{4} ({\sin 4 t|}_{0} = 0 . \end{array}$
$∴ g (x + π) = g (x) - g (π)$ also.

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