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`