How do you use the properties of integrals to verify the inequality \[\int{\dfrac{\sin \left( x \right)}{x}}\] from \[\dfrac{\pi }{4}\] to \[\dfrac{\pi }{2}\] is less than or equal to \[\dfrac{\sqrt{2}}{2}\]?
Answer
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Hint: In order to find the solution of the given question that is to verify that the inequality \[\int{\dfrac{\sin \left( x \right)}{x}}\] from \[\dfrac{\pi }{4}\] to \[\dfrac{\pi }{2}\] is less than or equal to \[\dfrac{\sqrt{2}}{2}\] use Maclaurin series of sine and substitute in the given integral and divide it by \[x\] then solve the integral and apply the given value of limits of the integral.
Complete step by step answer:
According to the question, we have show that the inequality \[\int{\dfrac{\sin \left( x \right)}{x}}\] from \[\dfrac{\pi }{4}\] to \[\dfrac{\pi }{2}\] is less than or equal to \[\dfrac{\sqrt{2}}{2}\], in other words we can say that:
\[\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx\le \dfrac{\sqrt{2}}{2}\]
We should start by noting that the integral of \[\dfrac{\sin \left( x \right)}{x}\] is not defined by real functions. Therefore, we will have to use the maclaurin series.
The maclaurin series for sine is known to be as follows:
\[\sin \left( x \right)=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{\left( n-1 \right)}}{{x}^{\left( 2n-1 \right)}}}{\left( 2n-1 \right)!}}=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+...\]
To determine the maclaurin series for \[\dfrac{\sin \left( x \right)}{x}\], we must divide each term by \[x\] from the above expression, we get:
\[\Rightarrow \dfrac{\sin \left( x \right)}{x}=\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{\left( n-1 \right)}}{{x}^{\left( 2n \right)}}}{\left( 2n-1 \right)!}}=1-\dfrac{{{x}^{2}}}{3!}+\dfrac{{{x}^{4}}}{5!}+...\]
Now to determine the value of \[\int{\dfrac{\sin \left( x \right)}{x}}dx\], we must integrate term by term (with respect to \[x\]) we get:
\[\Rightarrow \int{\dfrac{\sin \left( x \right)}{x}dx=}\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{\left( n-1 \right)}}{{x}^{\left( 2n+1 \right)}}}{\left( 2n+1 \right)\left( 2n-1 \right)!}}=x-\dfrac{{{x}^{3}}}{3\left( 3! \right)}+\dfrac{{{x}^{5}}}{5\left( 5! \right)}+...+C\]
As for the definite integral \[\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx\], we simply use the second fundamental theorem of calculus to evaluate it as follows:
\[\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx=\left[ x-\dfrac{{{x}^{3}}}{3\left( 3! \right)}+\dfrac{{{x}^{5}}}{5\left( 5! \right)}+...+C \right]_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}\]
\[\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx=\dfrac{\pi }{2}-\dfrac{{{\left( \dfrac{\pi }{2} \right)}^{3}}}{18}+\dfrac{{{\left( \dfrac{\pi }{2} \right)}^{5}}}{600}-\left( \dfrac{\pi }{4}-\dfrac{{{\left( \dfrac{\pi }{4} \right)}^{3}}}{18}+\dfrac{{{\left( \dfrac{\pi }{4} \right)}^{5}}}{600} \right)\]
After computing the sum of the first few terms, we get \[s=0.61\].
The integral converges to this value, because the more terms you add the less the decimals change (the actual value of the integral computed by calculator gives \[0.611786287085\] which is less than \[\dfrac{\sqrt{2}}{2}=0.707\].
\[\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx\le \dfrac{\sqrt{2}}{2}\]
The integral will satisfy the inequality as long as you use more than one term of the maclaurin series in the approximation.
Therefore, we can have proved that \[\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx\le \dfrac{\sqrt{2}}{2}\].
Note:
Students can go wrong by applying the wrong formula of maclaurin series of sine that is some student use \[\sin \left( x \right)=x+\dfrac{{{x}^{3}}}{3!}-\dfrac{{{x}^{5}}}{5!}+...\] which is completely wrong and leads to the wrong answer. It’s important to remember here that correct formula of maclaurin series of sine is \[\sin \left( x \right)=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{\left( n-1 \right)}}{{x}^{\left( 2n-1 \right)}}}{\left( 2n-1 \right)!}}=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+...\].
Complete step by step answer:
According to the question, we have show that the inequality \[\int{\dfrac{\sin \left( x \right)}{x}}\] from \[\dfrac{\pi }{4}\] to \[\dfrac{\pi }{2}\] is less than or equal to \[\dfrac{\sqrt{2}}{2}\], in other words we can say that:
\[\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx\le \dfrac{\sqrt{2}}{2}\]
We should start by noting that the integral of \[\dfrac{\sin \left( x \right)}{x}\] is not defined by real functions. Therefore, we will have to use the maclaurin series.
The maclaurin series for sine is known to be as follows:
\[\sin \left( x \right)=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{\left( n-1 \right)}}{{x}^{\left( 2n-1 \right)}}}{\left( 2n-1 \right)!}}=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+...\]
To determine the maclaurin series for \[\dfrac{\sin \left( x \right)}{x}\], we must divide each term by \[x\] from the above expression, we get:
\[\Rightarrow \dfrac{\sin \left( x \right)}{x}=\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{\left( n-1 \right)}}{{x}^{\left( 2n \right)}}}{\left( 2n-1 \right)!}}=1-\dfrac{{{x}^{2}}}{3!}+\dfrac{{{x}^{4}}}{5!}+...\]
Now to determine the value of \[\int{\dfrac{\sin \left( x \right)}{x}}dx\], we must integrate term by term (with respect to \[x\]) we get:
\[\Rightarrow \int{\dfrac{\sin \left( x \right)}{x}dx=}\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{\left( n-1 \right)}}{{x}^{\left( 2n+1 \right)}}}{\left( 2n+1 \right)\left( 2n-1 \right)!}}=x-\dfrac{{{x}^{3}}}{3\left( 3! \right)}+\dfrac{{{x}^{5}}}{5\left( 5! \right)}+...+C\]
As for the definite integral \[\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx\], we simply use the second fundamental theorem of calculus to evaluate it as follows:
\[\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx=\left[ x-\dfrac{{{x}^{3}}}{3\left( 3! \right)}+\dfrac{{{x}^{5}}}{5\left( 5! \right)}+...+C \right]_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}\]
\[\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx=\dfrac{\pi }{2}-\dfrac{{{\left( \dfrac{\pi }{2} \right)}^{3}}}{18}+\dfrac{{{\left( \dfrac{\pi }{2} \right)}^{5}}}{600}-\left( \dfrac{\pi }{4}-\dfrac{{{\left( \dfrac{\pi }{4} \right)}^{3}}}{18}+\dfrac{{{\left( \dfrac{\pi }{4} \right)}^{5}}}{600} \right)\]
After computing the sum of the first few terms, we get \[s=0.61\].
The integral converges to this value, because the more terms you add the less the decimals change (the actual value of the integral computed by calculator gives \[0.611786287085\] which is less than \[\dfrac{\sqrt{2}}{2}=0.707\].
\[\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx\le \dfrac{\sqrt{2}}{2}\]
The integral will satisfy the inequality as long as you use more than one term of the maclaurin series in the approximation.
Therefore, we can have proved that \[\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx\le \dfrac{\sqrt{2}}{2}\].
Note:
Students can go wrong by applying the wrong formula of maclaurin series of sine that is some student use \[\sin \left( x \right)=x+\dfrac{{{x}^{3}}}{3!}-\dfrac{{{x}^{5}}}{5!}+...\] which is completely wrong and leads to the wrong answer. It’s important to remember here that correct formula of maclaurin series of sine is \[\sin \left( x \right)=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{\left( n-1 \right)}}{{x}^{\left( 2n-1 \right)}}}{\left( 2n-1 \right)!}}=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+...\].
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