Use series to approximate the integral from 0 to 1 of [(1 - cos(2x))/ (x^2) ] dx

Question

Use series to approximate the integral from 0 to 1 of [(1 - cos(2x))/ (x^2) ] dx -----Check on calculator to make sure your error is less than .01. If error is not less than 0.01 or 1/100 take more terms in your series until error is less than 1/100. --------
Please show all steps in chronological order. First rating of five stars given to the person who shows the most detailed and correct answer. Thank you. Use series to approximate the integral from 0 to 1 of [(1 - cos(2x))/ (x^2) ] dx -----Check on calculator to make sure your error is less than .01. If error is not less than 0.01 or 1/100 take more terms in your series until error is less than 1/100. --------
Please show all steps in chronological order. First rating of five stars given to the person who shows the most detailed and correct answer. Thank you.

Explanation / Answer

Start with cos t = ?(n = 0 to ?) (-1)^n t^(2n) / (2n)!.

Let t = x^3:
cos(x^3) = ?(n = 0 to ?) (-1)^n x^(6n) / (2n)!

Multiply both sides by x:
x cos(x^3) = ?(n = 0 to ?) (-1)^n x^(6n+1) / (2n)!

Therefore, integrating both sides (term by term) between x = 0 and 1 yields
?(x = 0 to 1) x cos(x^3) = ?(n = 0 to ?) (-1)^n/[(6n+2) (2n)!].
-------------------------
Since the resulting series is alternating, the error after n terms is no bigger than the error of the following term.

For overkill, we insist that
1 / [(6(n+1) + 2) (2(n+1))!] < 10^(-4)
==> (6n+8) (2n+3)! > 10000
==> n ? 2 (by trial and error).

So, ?(x = 0 to 1) x cos(x^3)
? ?(n = 0 to 2) (-1)^n/[(6n+2) (2n)!]
? 0.440 (to three decimal places)

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.