How do you find the area under the graph of $f (x) = cos (x)$ $f(x)=cos(x)$ on the interval $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$ ?

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How do you find the area under the graph of $f (x) = cos (x)$ $f(x)=cos(x)$ on the interval $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$ ?

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Answer 1 · 2014-09-17T23:04:22

This is an integration problem. We will find the area under the curve $cos (x)$ $cos(x)$ over the interval $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$ . This is inclusive because of the square brackets.

On the unit circle remember that the positive side of y-axis corresponds to $\frac{π}{2}$ $pi/2$ and a coordinate of $(0, 1) .$ $(0,1).$ The y-coordinate corresponds to $1 .$ $1.$

On the unit circle remember that the negative side of y-axis corresponds to $- \frac{π}{2}$ $-pi/2$ and a coordinate of $(0, - 1) .$ $(0,-1).$ The y-coordinate corresponds to $- 1 .$ $-1.$

$\int_{- \frac{π}{2}}^{\frac{π}{2}} cos (x) d x$ $int_(-pi/2)^(pi/2)cos(x) dx$

$= {[sin (x)]}_{- \frac{π}{2}}^{\frac{π}{2}} = [sin (\frac{π}{2}) - sin (- \frac{π}{2})] = 1 - (- 1) = 2$ $=[sin(x)]_(-pi/2)^(pi/2)=[sin(pi/2)-sin(-pi/2)]=1-(-1)=2$

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How do you find the area under the graph of $f (x) = cos (x)$ $f(x)=cos(x)$ on the interval $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$ ?

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