Let the area of the region enclosed by the curve y = min{sin x, cos x} and the x- axis between x = –π to x = π be A. Then A2 is equal to ___________

Engineering

Mathematics

Introduction to Logarithm

Question

Let the area of the region enclosed by the curve y = min{sin x, cos x} and the x- axis between x = –π to x = π be A. Then A² is equal to ____________ .

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$f (x) = \{\begin{cases} cosx : - π < x < - \frac{3 π}{4} \\ sinx : - \frac{3 π}{4} \leq x \leq \frac{π}{4} \\ cosx : - \frac{π}{4} < π \leq π \end{cases}$

$A = \int_{- π}^{- 3 π / 4} - cosxdx + \int_{\frac{- 3 π}{4}}^{0} - sinxdx + \int_{0}^{π / 4} sinxdx + \int_{\frac{π}{4}}^{π / 2} cosxdx + \int_{\frac{π}{2}}^{π} - cosxdx$ $A = (- sinx)_{- π}^{- 3 π / 4} + (cosx)_{- 3 π / 4}^{0} + (- cosx)_{0}^{π / 4} + (sinx)_{π / 4}^{π / 2} + (- sinx)_{π / 2}^{π}$

$= (\frac{1}{\sqrt{2}} - 0) + (1 + \frac{1}{\sqrt{2}}) + (- \frac{1}{\sqrt{2}} + 1) + (1 - \frac{1}{\sqrt{2}}) + (- 0 + 1) = 4$

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