Volumes of Solids

Method: Max-Min Inequality Rule for Definite Integrals

Step 1: Determine the maximum y-value , f max , and the minimum y-value , f min , for your given equation, f ( x ) , on your given x-interval , [ a , b ].

You will not need to do a 1 st Derivative Test to find these values. Instead pause for a moment and consider two situations.

          Most often the equation, f ( x ) , you are given to work with is always increasing, or always decreasing on your given interval, [ a , b ]. This tells you that your minimum and maximum y-values will occur at the end points of your interval, [ a , b ].

          Another common situation is that your equation is a trig function like sine or cosine, which always bounce between a min = -1 and a max = 1.

Step 2: Multiply your f min value by ( b a ) to find the “lower bound” = f min ( b a ) .

This value is the smallest value that your definite integral, the exact net area,could be.

Step 3: Multiply your f max value by ( b a ) to find the “upper bound” = f max ( b a ) .

This value is the largest value that your definite integral, the exact net area,could be.

Step 4: Combine your results to show the interval (range between your “lower bound” and “upper bound”) that your definite integral must live between.

f min ( b a ) a b f ( x )   dx f max ( b a )

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