Exams Joint Entrance Examination Mathematics Area Under Curves Question 64 Question 64 of 230 Medium NUMERICAL
Let the area of the region enclosed by the curve \[y=\min \{\sin x, \cos x\}\] and the x x x axis between \[x=-\pi\] to \[x=\pi\] be A A A . Then A 2 A^2 A 2 is equal to \_\_\_\_\_\_\_\_\_\_.
Answer: 16
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y = f ( x ) = m i n sin x , cos x y=f(x)=min{\sin x,\cos x} y = f ( x ) = min sin x , cos x
A = − ∫ − π − 3 π 4 cos x d x − ∫ − 3 π 4 0 sin x d x + ∫ 0 π 4 sin x d x + ∫ π 4 π 2 cos x d x − ∫ π 2 π cos x d x A = 4 A 2 = 16 \begin{gathered} A=-\int _{-\pi }^{\frac{-3\pi }{4}}\cos xdx-\int _{\frac{-3\pi }{4}}^{0}\sin xdx+\int _{0}^{\frac{\pi }{4}}\sin xdx+\int _{\frac{\pi }{4}}^{\frac{\pi }{2}}\cos xdx \\ -\int _{\frac{\pi }{2}}^{\pi }\cos xdx \\ A=4 \\ {A}^{2}=16 \end{gathered} A = − ∫ − π 4 − 3 π cos x d x − ∫ 4 − 3 π 0 sin x d x + ∫ 0 4 π sin x d x + ∫ 4 π 2 π cos x d x − ∫ 2 π π cos x d x A = 4 A 2 = 16 Answer the question to unlock