Question on Taylor Polynomials Can someone please tell me how to go about this question? It's got me confused. 

Using the Taylor polynomial $P_{3}(x)$ of $e^x$, find a Taylor polynomial $P_{8}(x)$ for $x^{2}e^{-x^{2}}$. Use that to estimate the following integral: $$\int_{0}^{1}x^{2}e^{-x^{2}}dx.$$

 A: By Taylor,
$$P_3(x)=1+x+\frac{x^2}2+\frac{x^3}6$$ so that
$$x^2P_3(-x^2)=x^2-x^4+\frac{x^6}2-\frac{x^8}6.$$
As 
$$\int_0^1 x^kdx=\frac1{k+1},$$ the final answer is
$$\frac13-\frac15+\frac1{14}-\frac1{54}.$$

It is an easy matter to see that the general term is
$$\frac{(-1)^k}{(2k+3)k!}.$$
I couldn't resist to show the approximations to the true function (in blue) from polynomials $\color{green}{P_3}$, $P_4$ and $\color{magenta}{P_5}$.

A: Well first off you'd start off with the definition of a taylor polynomial, and we'll pick ours at a=0. 
$$f(x)=f(a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2!}+f'''(a)\frac{(x-a)^3}{3!}...$$
Now let's figure out the things we need to plug in.
$$\left(e^{x}\right)^{n}=e^{x}$$
where n is the nth derivative.
and
$$e^{0}=1$$
Plug it in what we found and you will get:
$$f(x)=1+x+\frac{x^2}{2}...$$
Now since we want $P_3$ we'll take up to and and including the term of degree 3.
Now to find a composition (i.e. $e^{-x^2})$ all you need to do is plug in -$-x^2$ into the polynomial we found above.
To compute the integral, plug in the polynomial we found with  $(-x^2)$ plugged in, multiply by $x^2$ and integrate!
........................
A: $$
P_3(x) = 1 + x + x^2/2 + x^3/6
$$
Note that $P_8(x) = x^2 P_3(-x^2)$
$$
P_8(x) = x^2 \left( 1 -x^2 + x^4/2 -x^6/6 \right) = x^2 - x^4 + x^6/2 - x^8/6
$$
Now you can simply integrate $P_8(x)$ using standard anti-derivative of polynomials !
A: $e^x = \sum \frac {x^n}{n!}\\
x^2 e^{-x^2} = \sum \frac {(-1^n)x^{2n+2}}{n!}\\
\int x^2 e^{-x^2} = \sum \frac {(-1^n)x^{2n+3}}{(2n+3)n!}$
