The key concept is to use information about derivatives to design the polynomial.
At x = 0, we find that f(0) = 1, f '(0) =1, and since f '(x) = f(x) we have that f ⁽ⁿ⁾(0) = 1 for all natural numbers n. We now consider the key result, typical of the Taylor's theory approach to estimation using polynomials based on derivative information.

THEOREM:
If   `P_n(x) = 1 + x + {x^2}/2 + {x^3}/{2*3) + {x^4}/{4!} + {x^5}/{5!} +...+ {x^n}/{n!} ` then P_n(x) is the unique polynomial of degree n so that

and P_n(b) is approximately equal to e^b.
In fact, if we let the error in this estimation be denoted by R_n = e^b - P_n(b) [ so e^b = P_n(b) + R_n ]
then for some c between 0 and b

To illustrate the key ideas of Taylor's theory, we start by estimating the number e with a polynomial. 

Estimating e: We apply the theorem using n = 5 to estimate the value of e [= e¹].
First,  ` P_5(x) = 1 + x + {x^2}/2 + {x^3}/{2*3) + {x^4}/{4!} + {x^5}/{5!} = 1 + x + {x^2}/2 + {x^3}/6 + {x^4}/24 + {x^5}/120 `.
So according to the theorem, e is approximately equal to 

and R₅ = e - P₅(1) where `R_5 = e^c  {1^6}/{6!} = e^c 1/720 ` for some c between 0 and 1. 

Thus, e is approximately 2.716667 and the error in using this estimate is R₅. Since c is between 0 and 1, e^c is between 1 and e, so we can deduce that R₅, the difference between e and the estimate, P₅(1), is no greater than e/720. Using estimates of e from your earlier work we know that e < 3, so our errror R₅ is no larger than `3/720 = 1/240`.

A more accurate estimate can be obtained by using a larger value for n. Try using n=6 and 7 to see the improvement. [This can be done progressively using a spreadsheet which you can examine now or later.]

Solution: For each t between 0 and 1, -t² is between -1 and 0. Let x = -t². By our theorem with n = 4, for each x between -1 and 0 we have

Since c < 0 , 0 < e^c < 1, and since x < 0 we have that `0 > R_4(x) > {x^5}/120`.

Substituting -t² for x ,we see that

Now since `0 > R_4(-t^2) > {(-t^2)^5}/120 = {-t^10}/120` we have `int_0^1 P_4(-t^2) dt > int_0^1 e^{-t^2} dt > int_0^1 P_4(-t^2) - {t^10}/120 dt`.

By evaluating these integrals we obtain

Therefore, 

and this is an overestimate by no more than `1/1320`.

Comment: We have been able to estimate a definite integral involving the exponential function by using a Taylor polynomial of degree 4. It should be apparent that by using a higher degree for the estimating polynomial, the error term will become smaller and we will obtain a more precise estimate. The systematic pattern in these polynomials should allow you to find more precise estimates for `int_0^1 e^{-x^2} dx` without much difficulty.

Proof of Proposition:  We'll begin our proof by finding P_n'(x).

`P_n'(x) = 1 + x + {x^2}/2  + {x^3}/{2*3} + {x^4}/{4!} + ... +{nx^{n-1}}/{n!} `
`= 1 + x + {x^2}/2 + {x^3}/{2*3} + {x^4}/{4!} + ... + {x^{n-1}}/{(n-1)!} = P_{n-1}(x)`

From this it follows easily that ` P_n(0) = P_n '(0) = P_n''(0) =...= P_n^{(n)}(0) = 1`.
Now suppose that b is not  0 and let R_n = e^b - P_n(b).
We need only justify the formula for evaluating R_n. For convenience we'll write R = R_n and let

Furthermore `g'(t) = e^t*P_n(b-t) - e^t*P_n'(b-t) - R* {(n+1)(b-t)^n}/{b^{n+1}}`
`= e^t* [P_n(b-t) - P_n'(b-t)] - R*{(n+1)(b-t)^n}/{b^{n + 1}}`
 
But `P_n(x) - P_n'(x) = {x^n}/{n!}, so 
`g'(t) = e^t*{(b-t)^n}/{n!} - R* {(n+1)(b-t)^n}/{b^{n+1}}`.

Now we apply the Mean Value (or Rolle's) Theorem to the function g, we can say that there is a number c between 0 and b where g'(c)=0. Thus

`0 = g'(c) = e^c*{(b-t)^n}/{n!} - R* {(n+1)(b-t)^n}/{b^{n+1}}` 
  and 

`e^t*{(b-t)^n}/{n!} = R* {(n+1)(b-t)^n}/{b^{n+1}}`.
Solving this last equation for R gives `R_n = R = e^c {b^{n+1}}/{(n+1)!}.