The area of the region enclosed by the curve $f (x) = \max {sinx, cosx}, - π \leq x \leq π$ and the x-axis is

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$2 \sqrt{2} (\sqrt{2} + 1)$

$4 (\sqrt{2})$

$2 (\sqrt{2} + 1)$

answer is B.

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Detailed Solution

$A r e a = |\int_{- π}^{- 3 π / 4} \sin x d x| + |\int_{- 3 π / 4}^{- π / 2} \cos x d x| + |\int_{- π / 2}^{π / 4} \cos x d x| + |\int_{π / 4}^{π} \sin x d x| = \int_{- π}^{- 3 π / 4} - sinxdx + \int_{\frac{- 3 π}{4}}^{- π / 2} - cosxdx + \int_{- \frac{π}{2}}^{\frac{π}{4}} cosxdx + \int_{π / 4}^{π} sinxdx = 4$
$\begin{array}{l} = |\frac{1}{\sqrt{2}} - 1| + |1 + \frac{1}{\sqrt{2}}| + |- 1 + \frac{1}{\sqrt{2}}| + |\frac{1}{\sqrt{2}} + 1| \\ = 2 + \frac{2}{\sqrt{2}} + 2 - \frac{2}{\sqrt{2}} = 4 \end{array}$