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

May 5, 2018

$F \left(x\right) = x - {x}^{5}$

Explanation:

Using Fundamental Theorem of Calculus:

$F ' \left(x\right) = {e}^{- 5 {x}^{4}}$

By definition:

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

$\implies {e}^{- 5 {x}^{4}} = 1 - 5 {x}^{4} + m a t h \boldsymbol{O} \left({x}^{8}\right) = F ' \left(x\right)$

Integrating:

• $F \left(x\right) = C + x - {x}^{5} + m a t h \boldsymbol{O} \left({x}^{9}\right)$

And:

• $F \left(0\right) = 0 \implies C = 0$

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.