How do you use the properties of integrals to verify the inequality #intsinx/x# from pi/4 to pi/2 is less than or equal to #sqrt(2)/2#?

Question

Emily Ammerman · Accepted Answer

We wish to show that

#int_(pi/4)^(pi/2) sinx/x dx≤ sqrt(2)/2#

We should start by noting that the integral of #sinx/x# is not defined by real functions. Therefore we will have to use maclaurin series.

The maclaurin series for sine is known to be #sinx = sum_(n = 1)^oo ((-1)^(n - 1)x^(2n - 1))/((2n - 1)!) = x - x^3/(3!) + x^5/(5!) + ... #

To determine the maclaurin series for #sinx/x#, we must divide each term by #x#.

#sinx/x = sum_(n = 0)^oo ((-1)^(n - 1)x^(2n))/((2n -1)!) = 1 - x^2/(3!) + x^4/(5!) + ...#

To determine the value of #int sinx/x dx#, we must integrate term by term (with respect to #x#).

#int sinx/xdx = sum_(n = 0)^oo ((-1)^(n - 1)x^(2n + 1))/((2n + 1)(2n - 1)!) = x - x^3/(3(3!)) + x^5/(5(5!)) +... + C#

As for the definite integral #int_(pi/4)^(pi/2) sinx/x dx#, we simply use the second fundamental theorem of calculus to evaluate.

#int_(pi/4)^(pi/2) sinx/x dx = pi/2 - (pi/2)^3/18 + (pi/2)^5/600 - (pi/4 - (pi/4)^3/18 + (pi/4)^5/600)#

Even 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 #0.611786287085706#. This is less than #sqrt(2)/2 = 0.707#. The integral will satisfy the inequality as long as you use more than one term of the maclaurin series in the approximation (which you would, because one term is extremely imprecise).

Hopefully this helps!

Paisley Johnson · Answer

undefined To verify the inequality $ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sin(x)}{x} \, dx \leq \frac{\sqrt{2}}{2} $, we can compare it with a known integral and utilize the properties of integrals.

1. Define $ f(x) = \frac{\sin(x)}{x} $.
2. Note that $ f(x) $ is continuous on the interval $ [\frac{\pi}{4}, \frac{\pi}{2}] $ except possibly at $ x = 0 $.
3. Since $ \sin(x) $ is non-negative on $ [\frac{\pi}{4}, \frac{\pi}{2}] $, we have $ 0 \leq f(x) $.
4. Observe that $ \frac{\sqrt{2}}{2} $ is the maximum value of $ f(x) $ on $ [\frac{\pi}{4}, \frac{\pi}{2}] $ since $ \frac{\sin(x)}{x} $ is decreasing on this interval.
5. Therefore, $ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sin(x)}{x} \, dx \leq \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sqrt{2}}{2} \, dx = \frac{\sqrt{2}}{2} \cdot \frac{\pi}{2} = \frac{\pi\sqrt{2}}{4} $.

Thus, we have verified that $ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sin(x)}{x} \, dx \leq \frac{\sqrt{2}}{2} $.