I need to Evaluate at X=.3 and .5 with precision of n=8 and I need to find the maximum error. I am very lost. Thank you.
$\displaystyle \int_{0}^{X} t e^-t^3$
i know that $\displaystyle e^x = \sum_{n=1}^\infty x^n/n! $ so then $\displaystyle t * \sum_{n=1}^\infty (-t^3)^n/n!$
Does that equal $\displaystyle \sum_{n=1}^\infty (-t^3)^{n+1}/n!$ or is it something else entirely? When I put it in my calculator I get something else entirely but someone else told me that it works??
Ok, first notice that
$\displaystyle \forall{t}\in\mathbb{R}~e^{t}=\sum_{n=0}^{\infty}\ frac{t^n}{n!}$
So
$\displaystyle \begin{aligned}\forall{-t^3}\in\mathbb{R}\implies{t}\in\mathbb{R}~te^{-t^3}&=t\sum_{n=0}^{\infty}\frac{\left(-t^3\right)^n}{n!}\\
&=t\sum_{n=0}^{\infty}\frac{(-1)^nt^{3n}}{n!}\\
&=\sum_{n=0}^{\infty}\frac{(-1)^nt^{3n+1}}{n!}
\end{aligned}$
Now notice that a power series is uniformly convergent, thus integrable on its interval of convergence so
$\displaystyle \begin{aligned}\int_0^{x}te^{-t^3}dy&=\int_0^x\sum_{n=0}^{\infty}\frac{(-1)^nt^{3n+1}}{n!}dt\\
&=\sum_{n=0}^{\infty}\int_0^1\frac{(-1)^nt^{3n+1}}{n!}dt\\
&=\sum_{n=0}^{\infty}\frac{(-1)^nx^{3n+2}}{(3n+2)n!}\end{aligned}$
Now input the appropriate values and note that if $\displaystyle S=\sum_{n=0}^{\infty}a_n$ and $\displaystyle S_N=\sum_{n=0}^{N}a_n$ that
$\displaystyle R_N=\left|S-S_N\right|\leqslant{a_{N+1}}$
Or in other words the error is less than or equal to the first neglected term. So lets say your error is $\displaystyle e$, then you need to find $\displaystyle N$ such that $\displaystyle a_{N+1}\leqslant{e}$ and that will be the correct number of terms. But remember that since $\displaystyle N\in\mathbb{N}$ if you calcluate the correct $\displaystyle N$ to be lets say $\displaystyle \varphi$ you really want $\displaystyle \lceil{\varphi}\rceil$.