int_0^1 e^(-x^2) dx Use a power series to approximate the value of the integral with an error of less than 0.0001.

From the table of power series, we have:
e^x = sum_(n=0)^oo x^n/n!
     = 1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+x^5/(5!)+ ...
 To apply this on the given integral int_0^1 e^(-x^2)dx ,
we replace the "x " with "-x^2 ".
e^(-x^2)= sum_(n=0)^oo (-x^2)^n/(n!)  
          =sum_(n=0)^oo ((-1)^n*x^(2n))/(n!)
          = 1/(0!) -x^2/(1!)+x^4/(2!) - x^6/(3!) +x^8/4!-x^(10)/(5!)+x^(12)/(6!) -...
          = 1 -x^2 +x^4/2-x^6/6 +x^8/24-x^(10)/120+x^(12)/(6!)- ...
The integral becomes:
int_0^1 e^(-x^2)dx =int_0^1 [1 -x^2 +x^4/2-x^6/6 +x^8/24-x^(10)/120+x^(12)/720-...]dx
To determine the indefinite integral, we integrate each term using Power Rule for integration: int x^ndx =x^(n+1)/(n+1) .
int_0^1 [1 -x^2 +x^4/2-x^6/6 +x^8/24-x^(10)/120+x^(12)/720-...]dx
=[x-x^3/3 +x^5/(2*5)-x^7/(6*7) +x^9/(24*9)-x^(11)/(120*11)+x^(13)/(720*13)-...]|_0^1
=[x-x^3/3 +x^5/10-x^7/42+x^9/216-x^(11)/1320+x^(13)/9360-...]|_0^1
Apply definite integral formula: F(x)|_a^b = F(b) - F(a) .
F(1) = 1-1^3/3 +1^5/10-1^7/42+1^9/216-1^(11)/1320+1^(13)/9360- ...
         = 1 -1/3 +1/10-1/42 +1/216-1/1320+1/9360- ...
F(0) = 0-0^3/3 +0^5/10-0^7/42+0^9/216-0^(11)/1320+0^(13)/9360- ...
          = 0 -0 +0 -0 +0-0+0- ...      
All the terms are 0 then F(0)= 0 .
We can stop at 7th term (1/9360 ~~0.0001068) since we only need error less than 0.0001.
Then,
 F(1)-F(0)=[1 -1/3 +1/10-1/42 +1/216-1/1320+1/9360] -[0]
                      = 0.7468360343
Thus, the approximation of the integral will be:
int _0^1 e^(-x^2)dx ~~0.7468