This has to do with Maclaurin series?

Question

Assume that sin(x) equals its Maclaurin series for all x.

Use the Maclaurin series for sin(4 x^2) to evaluate the integral

\int_0^{0.62} \sin(4 x^2) \ dx

. Your answer will be an infinite series. Use the first two terms to estimate its value if anyone has an answer that would be great

Answer 1

Recall that the Maclaurin (Taylor) series for sin x is

sin x = x - x^3/3! + x^5/4! - x^7/7! +- ...

so

sin (4x^2) = 4x^2 - (4^3/3!) x^6 + (4^5/5!) x^10 - (4^7/7!) x^14 +- ...

The integral can be evaluated term by term

int_0^0.62 \sin(4 x^2) \ dx =

  (4/3)*0.62^3 - (4^3/(7*3!)) 0.62^7 + (4^5/(11*5!)) 0.62^11

       - (4^7/(15*7!)) 0.62^15 +-...

Using the first 2 terms as an approximation,

int_0^0.62 \sin(4 x^2) \ dx = 0.317771 - 0.0536627 ...

  = 0.2641...

The error in this approximation is bounded by the size of the third term (because the series is an alternating, decreasing series), which is

(4^5/(11*5!)) 0.62^11 = 0.00404...

so two terms does not provide a particularly accuracte approximation.