Prove that if f is Riemann Integrable on [a,b] and m<=f(x)<=M for all x in the interval then m(b-a)<=integral from a to b f<=M(b-a)
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Draw three graphs. (a) Graph y= f(x), (b) graph y= M, (c) graph y= m. You are told that m<= f(x)<= M for all x from a to b. Remember that the integral of a function is "the area under the graph".
Originally Posted by HallsofIvy Remember that the integral of a function is "the area under the graph". I did not know that? $\int_0^{2\pi } {\sin (x)} = 0$