recall that
sin(x) = x - x^3/3! + x^5/5! - x^7/7! for the first 4 terms
so replacing x with x^2
sin(x^2 = x^2 - x^6/6 + x^10/120 + x^14/5040
ʃsin(x^2) = ʃ(x^2 - x^6/6 + x^10/120 + x^14/5040) dx from 0 to 1
= [(1/3)x^3 - (1/42)x^7 + (1/1320)x^11 - (1/75600)x^15] from 0 to 1
= (1/3 - 1/42 + 1/1320 - 1/75600) - 0
= .310268
confirmation
https://www.wolframalpha.com/input/?i=%CA%83+sin(x%5E2)+dx+from+0+to+1
So my answer using 4 terms is correct to 6 decimal places
for c) take one less term and see if the answer changes correct to 3 decimals.
The prompt for this question is f(x) =sin(x^2)
A)Write the first four terms of the Maclaurin series for f(x)
B)Use the Maclaurin series found in Part A to approximate the integral from 0 to 1 of sin(x^2) dx
C)How many terms are needed to find the value of the integral given in Part B, correct to three decimal places? What is that value?
I have no idea what to do for this whole problem, please write out the steps
3 answers
f(x) = f(0) + x f'(0) + (x^2/2)f''(0) +(x^3/6)f'''(0) ......
f(0) = sin(x^2) = 0
f'(0) = 2xcosx^2 = 2*0 = 0
f''(0) = -2x(2x)sinx^2 +2cosx^2 = 2
f'''(0) = -8x^3cosx^2 -8xsinx^2-4xcosx^2
= -8x^3
so
f(x) = 2(x^2/2) -8 x^3(x^3/6)
= x^2 -(4/3)x^6
now integrate
x^3/3 - (4/3)x^7/7
= x^3/3 - 4x^7/21
CHECK MY ARITHMETIC
f(0) = sin(x^2) = 0
f'(0) = 2xcosx^2 = 2*0 = 0
f''(0) = -2x(2x)sinx^2 +2cosx^2 = 2
f'''(0) = -8x^3cosx^2 -8xsinx^2-4xcosx^2
= -8x^3
so
f(x) = 2(x^2/2) -8 x^3(x^3/6)
= x^2 -(4/3)x^6
now integrate
x^3/3 - (4/3)x^7/7
= x^3/3 - 4x^7/21
CHECK MY ARITHMETIC
Okay I changed C) and used one less term and got 0.31028 as the answer. What does that mean?
0 answers