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The area bounded by the curves y = \sin x and y = \cos x between two consecutive points of their intersection, is (in sq. units)

  • Option 1)

    \sqrt{2}

  • Option 2)

    2

  • Option 3)

    2\sqrt2

  • Option 4)

    2+\sqrt2

 

Answers (1)

best_answer

As we learnt 

 

Area between two curves -

 

\\*I\! \! f \: f\left ( x \right )\geqslant g\left ( x \right )\\* in[a,c)\: \: and \: \: g\left ( x \right ) \geqslant f\left ( x \right )\:in(c,b]\\* Then\: area = \\*\\* \int_{a}^{c}\left ( f\left ( x \right )-g\left ( x \right ) \right )dx+\int_{c}^{b}\left ( g\left ( x \right )-f\left ( x \right ) \right )dx

- wherein

 

 Two consecutive points of intersections of y = \sin x$ and y = \cos x$

can be taken as  x = \frac{\pi }{4}$ and x = \frac{5\pi }{4}$

\         Required area =

                                           \int\limits_{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}}^{{{5\pi } \mathord{\left/ {\vphantom {{5\pi } 4}} \right. \kern-\nulldelimiterspace} 4}} {\,\,\left( {\sin x - \cos x} \right)} \,dx = \left[ { - \cos x - \sin x_{}^{}} \right]_{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}}^{{{5\pi } \mathord{\left/ {\vphantom {{5\pi } 4}} \right. \kern-\nulldelimiterspace} 4}}

                                               =\frac{2}{{\sqrt 2 }} + \frac{2}{{\sqrt 2 }} = 2\sqrt 2sq. units.


Option 1)

\sqrt{2}

Option 2)

2

Option 3)

2\sqrt2

Option 4)

2+\sqrt2

Posted by

gaurav

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