# Question 4cacf

Apr 10, 2017

We have:

int_0^1 e^(-x^3) dx = sum_(n=0)^oo (-1)^n/(n!) 1/(3n+1)

and we can stop the sum at $n = 3$ to have an error less than $0.01$:

${\int}_{0}^{1} {e}^{- {x}^{3}} \mathrm{dx} \cong \frac{169}{210}$

#### Explanation:

We know that the Taylor series expansion of ${e}^{t}$ around $t = 0$ is:

e^t = sum_(n=0)^oo t^n/(n!)

Substituting $t = - {x}^{3}$:

e^(-x^3) = sum_(n=0)^oo (-1)^n x^(3n)/(n!)

and since this series has radius of convergence $R = \infty$ we can integrate term by term:

int_0^1 e^(-x^3) dx = sum_(n=0)^oo (-1)^n int_0^1x^(3n)/(n!)dx

int_0^1 e^(-x^3) dx = sum_(n=0)^oo (-1)^n/(n!) [x^(3n+1)/(3n+1)]_0^1

int_0^1 e^(-x^3) dx = sum_(n=0)^oo (-1)^n/(n!) 1/(3n+1)

If we stop the sum at the term $n = N$, Lagrange's theorem states that:

int_0^1 e^(-x^3) dx = sum_(n=0)^N (-1)^n/(n!) 1/(3n+1)+ (-1)^(N+1)/((N+1)!) xi^(3N+1)/(3N+1) with $0 \le \xi \le 1$

As $\xi \le 1 \implies {\xi}^{3 N + 1} \le 1$ we have that:

int_0^1 e^(-x^3) dx = sum_(n=0)^N (-1)^n/(n!) 1/(3n+1)+R_N

where:

abs(R_N) <= 1/((N+1)!)1/(3N+1)#

So:

$\left\mid {R}_{1} \right\mid \le \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}$

$\left\mid {R}_{2} \right\mid \le \frac{1}{6} \cdot \frac{1}{7} = \frac{1}{42}$

$\left\mid {R}_{3} \right\mid \le \frac{1}{24} \cdot \frac{1}{10} = \frac{1}{240} < 0.01$

So to have an error smaller than $0.01$ we need four terms:

${\int}_{0}^{1} {e}^{- {x}^{3}} \mathrm{dx} \cong 1 - \frac{1}{4} + \frac{1}{14} - \frac{1}{60} = \frac{420 - 105 + 30 - 7}{420} = \frac{338}{420} = \frac{169}{210}$