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?

1 Answer
Mar 24, 2018

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!