What is the net area between $f (x) = x - sin x$ $f(x) = x-sinx$ and the x-axis over $x \in [0, 3 π]$ $x in [0, 3pi ]$ ?

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Answer 1 · 2018-05-07T22:11:32

$int_0^(3π)(x-sinx)dx=((9π^2)/2-2) m^2$

$f(x)=x-sinx$ , $x$ $in$ $[0,3pi]$

$f(x)=0$ $<=>$ $x=sinx$ $<=>$ $(x=0)$

(Note: $|sinx|<=|x|$ , $AA$ $x$ $in$ $RR$ and the $=$ is true only for $x=0$ )

So when $x$ $in$ $[0,3pi]$ , $f(x)>=0$

Graphical help enter image source here

The area we are looking for since $f(x)>=0$ , $x$ $in$ $[0,3pi]$

is given by $int_0^(3π)(x-sinx)dx$ $=$

$int_0^(3π)xdx$ $- int_0^(3π)sinxdx$ $=$

$[x^2/2]_0^(3π)+[cosx]_0^(3π)$ $=$

$(9π^2)/2+cos(3π)-cos0$ $=$

$((9π^2)/2-2)$ $m^2$

What is the net area between f(x) = x-sinx f(x)=x−sinxf(x) = x-sinx and the x-axis over x in [0, 3pi ]x∈[0,3π]x in [0, 3pi ]?