Question 1189664
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If R is a region between the graphs of the function f(x)=sinx and g(x)=cosx over the interval [0,pie]
a) find the area of rigion R
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            In this post,  I will solve problem  (a),  ONLY.



<pre>
If you plot the graphs of the functions f(x) = sin(x) and g(x) = cos(x) over the interval [0,pi],

you will see that  g(x) >= f(x)  at 0 <= x <= {{{pi/4}}}  and  g(x) <= f(x)  at {{{pi/4}}} <= x <= {{{pi}}}.

Also, you can get it algebraically.



In any case, the area of the region R is the sum of two integrals


    area(R) = {{{int((cos(x)-sin(x)), dx, 0, pi/4)}}} + {{{int((sin(x)-cos(x)), dx, pi/4, pi)}}}.



First integral equals

    (sin(x) + cos(x))  from 0 to {{{pi/4}}},  which is  {{{2*(sqrt(2)/2)}}} - 1 = {{{sqrt(2)-1)}}}.



Second integral equals

    (-cos(x) - sin(x))  from {{{pi/4}}} to {{{pi}}},  which is  1 + {{{2*(sqrt(2)/2)}}} = {{{sqrt(2)+1}}}.


After adding the integral values, we get

    area(R) = {{{sqrt(2)-1)}}} + {{{sqrt(2)+1)}}} = {{{2*sqrt(2)}}} = 2.828427  (rounded).    <U>ANSWER</U>
</pre>

Part (a) is solved.