Method: Max-Min Inequality Rule for Definite Integrals

Step 1: Determine the maximum y-value , $f_{\mathit{max}}$, and the minimum y-value , $f_{\mathit{min}}$, for your given equation, $f\left(x\right)$, 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, $f\left(x\right)$, 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 $f_{\mathit{min}}$ value by ( b – a ) to find the “lower bound” $=f_{\mathit{min}}\bullet \left(b–a\right)$.

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

Step 3: Multiply your $f_{\mathit{max}}$ value by ( b – a ) to find the “upper bound” ${=f}_{\mathit{max}}\bullet \left(b–a\right)$.

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.

$f_{\mathit{min}}\bullet \left(b–a\right)\le \underset{a}{\overset{b}{\int }}f\left(x\right) \mathit{dx}\mathit{\le }{f}_{\mathit{max}}\bullet \left(b–a\right)$