Example 1: Max-Min Inequality for Definite Integrals

Show that the value of $\underset{0}{\overset{1}{\int }}\mathit{sin}\left(x^2\right)\mathit{dx}$ cannot possibly be 2.

Identifier: You are being given a definite integral and are not being asked at all about its true value, just to show that it could not be a specific value.

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 ].

Here your $f\left(x\right)=\mathit{sin}\left(x^2\right)$ which is a trig function that always oscillates between a minimum y-value of -1 , and a maximum y-value of 1 .

$f_{\mathit{min}}=–1$

$f_{\mathit{max}}=1$

Keep in mind in these situations that it does not matter what is happening inside the sine or cosine function. You know that they must always oscillate between -1 and 1.

Step 2: Multiply your $f_{\mathit{min}}$ value by ( b – a ) to find the “lower bound” $=f_{\mathit{min}}\bullet \left(b–a\right)$.

The interval [ a , b ]= [ 0 , 1 ] is provided by the bounds on your definite integral, $\underset{0}{\overset{1}{\int }}$.

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

$f_{\mathit{min}}=–1$

$\left(b–a\right)=\left(1–0\right)=1$

lower bound$=f_{\mathit{min}}\bullet \left(b–a\right)=\left(–1\right)\bullet \left(1\right)=–1$

lower bound = -1

$–1\le \underset{0}{\overset{1}{\int }}\mathit{sin}\left(x^2\right)\mathit{dx}$

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.

$f_{\mathit{max}}=1$

$\left(b–a\right)=\left(1–0\right)=1$

upper bound $=f_{\mathit{max}}\bullet \left(b–a\right)=\left(1\right)\bullet \left(1\right)=1$

upper bound = 1

$\underset{0}{\overset{1}{\int }}\mathit{sin}\left(x^2\right)\mathit{dx}\mathit{\le }1$

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

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

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

$–1\le \underset{0}{\overset{1}{\int }}\mathit{sin}\left(x^2\right)\mathit{dx}\mathit{\le }1$

Final Result:

Based upon the Max-Min Inequality Rule the definite integral $\underset{0}{\overset{1}{\int }}\mathit{sin}\left(x^2\right)\mathit{dx}$ has to have a value between -1 and 1, $–1\le \underset{0}{\overset{1}{\int }}\mathit{sin}\left(x^2\right)\mathit{dx}\mathit{\le }1$. This means that there is no way that the definite integral could be 2 because 2 is not between -1 and 1.

$\underset{0}{\overset{1}{\int }}\mathit{sin}\left(x^2\right)\mathit{dx}\ne 2$