Use Maclaurin series to evaluate the integral

Assume that (sinx) equals its Maclaurin series for all x.
Use the Maclaurin series for sin(8x^2) to evaluate the integral

a = 0 b = 0.65 sin(8x^2)dx


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

Remember sin x= x-x^3/3!+x^5/5!....
Int sin(8x^2) dx=
Int 8x^2-(8x^2)^3/3!...dx
8Int x^2-64(x^2)^3/3!...dx
8[x^3/3-64x^7/42]
Inserting limits
8[.65^3/3-32(.65^7)/21]
8/21[7(.65^3)-32(.65^7)]
8/21[.65^3][7-32(.65^4)]
Out with the calculator

The Maclaurin series for sin x ≈ x -x^3/3 + x^5/5 - x^7/7 + .......

Therefore:

sin(8x^2) ≈ (8x^2) - (8x^2)^3 / 3 + (8x^2)^5 / 5 - (8x^2)^7 / 7 + .......

sin(8x^2) ≈ 8x^2 - 512x^ 6 / 3 + (8^5 * x^10 / 5 - 8^7 * x^14 / 7 + .......

Now integrating the series:

∫ sin(8x^2)dx =

= ∫ 8x^2 - 512x^ 6 / 3 + (8^5 * x^10 / 5 - 8^7 * x^14 / 7 dx =

= 8x^3 / 3 - 256x^7 / 21 + 4096x^11 / 165 - 131072x^15 / 4725 (from 0 to o.65)

= 0.313618

The use of the calculator is essential

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