Question

Let `F(x)=int_0^x e^(-5t^4)dt`. Find the MacLaurin polynomial of degree 5 for F(x)?

Answer 1

`F(x)= x - x^5`

Explanation:

Using Fundamental Theorem of Calculus:

`F'(x) = e^(-5x^4)`

By definition:

`e^(z)= sum_{k=0}^{oo } z^k/(k!) =1+z + mathbb O(z^2)`

`implies e^(-5x^4)= 1 -5x^4 + mathbb O(x^8) = F'(x)`

Integrating:

• `F(x)= C + x - x^5 + mathbb O(x^9)`

And:

• `F(0) = 0 implies C = 0`

You can get the same result by evaluating a whole series of derivatives and calculating as the Maclaurin Series:

`f(x)=f(0)+f^'(0)x+(f^('')(0))/(2!)x^2+(f^((3))(0))/(3!)x^3+...+(f^((n))(0))/(n!)x^n+....`

But I would not recommend it.