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The area, enclosed by the curves $$y = \sin x + \cos x$$ and $$y = \left| {\cos x - \sin x} \right|$$ and the lines $$x = 0,x = {\pi \over 2}$$, is :

JEE · Math · previous-year question

  1. A.$$2\sqrt 2 (\sqrt 2 - 1)$$correct
  2. B.$$2(\sqrt 2 + 1)$$
  3. C.$$4(\sqrt 2 - 1)$$
  4. D.$$2\sqrt 2 (\sqrt 2 + 1)$$

Answer

A. $$2\sqrt 2 (\sqrt 2 - 1)$$

Explanation

$$A = \int_0^{{\pi \over 2}} {\left( {(\sin x + \cos x) - \left| {\cos x - \sin x} \right|} \right)\,dx} $$ $$A = \int_0^{{\pi \over 2}} {\left( {(\sin x + \cos x) - (\cos x - \sin x)} \right)\,dx} + \int_{{\pi \over 4}}^{{\pi \over 2}} {\left( {(\sin x + \cos x) - (\sin x - \cos x)} \right)\,dx} $$ $$A = 2\int_0^{{\pi \over 2}} {\sin x\,dx + 2\int_{{\pi \over 4}}^{{\pi \over 2}} {\cos x\,dx} } $$ $$A = - 2\left( {{1 \over {\sqrt 2 }} - 1} \right) + \left( {1 - {1 \over {\sqrt 2 }}} \right)$$ $$A = 4 - 2\sqrt 2 = 2\sqrt 2 (\sqrt 2 - 1)$$ Option (a)

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