We know that
{f_{avg}} = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)\,dx} \
Solution (a)
f(x)=9−2e4x+1 for the interval [2,6]
favg=6−212∫6(9−2e4x+1)dxfavg=492∫6(1)dx−422∫6(e4x+1)dxfavg=49[x]26−21[4e4x+1]26favg=49[6−2]−81[e4x+1]26favg=49[4]−81[e4(6)+1−e4(2)+1]favg=9−81[e25−e9]favg=9−81[e25−e9]favg=−9000611395
Then for f(c)=favg means
f(c)=favg9−2e4c+1=−90006113952e4c+1=9+90006113952e4c+1=9000611404e4c+1=45003057024c+1=ln(4500305702)4c+1=22.227411174c=21.22741117c=5.306852791
Solution (b)
f(x)=8−cos(4x) for the interval [0,4π]
favg=4π−010∫4π(8−cos(4x))dxfavg=4π10∫4π(8)dx+4π10∫4π(cos(4x))dxfavg=4π8[x]04π+4π1[41sin(4x)]04πfavg=π2[4π−0]+4π4[sin(4x)]04πfavg=8+π1[sin(44π)−sin(40)]favg=8+π1[0−0]favg=8
Now for
f(c)=favg8−cos(4c)=8cos(4c)=04c=2πc=2π
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