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Find the average rate of change of f(x)=1+sin x over interval [0,pi]
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Average rate of change of a function \(f\) over some interval \([a,b]\), provided that \(f\) is continuous over the interval, is \[\frac{f(b)-f(a)}{b-a}\]
you got this?
@SithsAndGiggles Same thing came to my head but what is a? it is 0?
Yes, \([a,b]=[0,\pi]\), so \(a=0\) and \(b=\pi\).
OK. Now I'm calm. Oh you forgot to add the limit as b approaches a.
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That's the case for instantaneous rate of change. We just want the average rate of change.
Oh lol I didn't read the question well.
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