First slide

Evaluation of definite integrals

Question

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

Easy

A

$g (x) + g (π)$
B

$g (x) - g (π)$
C

$g (x) g (π)$
D

$g (x) / g (π)$

Solution

$\begin{aligned} g (x + π) = \int_{0}^{x + π} \cos^{4} t d t \\ = \int_{0}^{π} \cos^{4} t d t + \int_{π}^{x + π} \cos^{4} t d t \\ = g (π) + I_{1} \end{aligned}$
In $I_{1},$ put $t = π + u,$ so that
$I_{1} = \int_{0}^{x} \cos^{4} (π + u) d u = \int_{0}^{x} \cos^{4} u d u = g (x) .$

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