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The area, enclosed by the curves y=sin x + cosx and y=|cosx-sinx| and the lines x=0, x =π/2, is |
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Answer» cosx > sinx, ∀x ∈ (0,\(\frac{\pi}{4}\)), and cosx < sinx, ∀x ∈ (\(\frac{\pi}{4}\),\(\frac{\pi}{2}\)) y1 = sinx + cosx y2 = |cosx - sinx| ⇒ Area = \(\int_0^{\pi/2}(y_1-y_2)dx\) = \(\int_0^{\pi/4}((sinx+cosx)\)\(-(cosx-sinx))dx\) + \(\int_{\pi/4}^{\pi/2}((sinx+cosx)\)\(-(sinx-cosx))dx\) = 4 - 2√2 |
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