Use the Maclaurin series for e^(-3x^4) to evaluate the integral

Use the Maclaurin series for e^(-3x^4) to evaluate the integral from 0 to 0.17 of e^(-3x^4) dx, use the first two terms.

Ok, I am having troubles with this one, I understand the formula: f(0)+f'(0)x/1!+f''(0)x^2/2!+...... But how exactly do I solve this problem?
Any help would be greatly appreciated.

Start with the basic exponential series
e^t = 1 + t + t^2/2! + ...

Let t = -3x^4:
e^(-3x^4) = 1 + (-3x^4) + (-3x^4)^2 / 2! + ...
..............= 1 - 3x^4 + (9/2)x^8 + ...

Integrating term by term yields
∫(x = 0 to 0.17) e^(-3x^4) dx
= ∫(x = 0 to 0.17) [1 - 3x^4 + (9/2)x^8 + ...] dx
= [x - (3/5)x^5 + (1/2)x^9 + ...] {for x = 0 to 0.17}
= 0.17 - (3/5)(0.17)^5 + (1/2) (0.17)^9 + ...

Using the first two terms,
∫(x = 0 to 0.17) e^(-3x^4) dx ≈ 0.17 - (3/5)(0.17)^5 ≈ 0.1699.

I hope this helps!