Maclaurin series for calc 2?

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 sin(4x^2)dx form 0-.62

Your answer will be an infinite series. Use the first two terms to estimate its value.

i need an answer please

Answer 1

The Maclaurin series for sin(x) is

∞

∑ (-1)^n x^(2n+1) / (2n+1)!

n=0

Just replace x by 4x² to get the Maclaurin series for sin(4x²):

∞

∑ (-1)^n (4x²)^(2n+1) / (2n+1)!

n=0

which simplifies to

∞

∑ (-1)^n [4^(2n+1)][x^(4n+2)] / (2n+1)!

n=0

The integral of sin(4x²) equals the integral of this sum. Assuming that interchanging summation and integration is legitimate here (which it is), the integral of sin(4x²) is given by

∞

∑ ∫ {(-1)^n [4^(2n+1)][x^(4n+2)] / (2n+1)!} dx

n=0

which equals

∞

∑ (-1)^n [4^(2n+1)][x^(4n+3)] /[(4n + 3) (2n+1)!]

n=0

For the definite integral from 0 to 0.62, the value is

∞

∑ (-1)^n [4^(2n+1)][0.62^(4n+3)] /[(4n + 3) (2n+1)!]

n=0

since the infinite sum, evaluated at 0, is 0.

Using only the first two terms.

0.62

∫ sin(4x²) dx ≈

0

0.62

∫ (4x² - (32x^6)/3) dx

0

= (4/3)x³ - (32/21) x^7 (evaluated at 0 and 0.62)

≈ 0.264