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

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marizi eNotes educator| Certified Educator

 From the table of power series, we have:

`cos(x) = sum_(n=0)^oo (-1)^nx^(2n)/((2n))!`

            `= 1-x^2/(2!)+x^4/(4!)-x^6/(6!)+` ...

 To apply this on the given integral `int_0^1 cos(x^2) dx` , we may replace the "`x` " with "`x^2` ".

`cos(x^2) = sum_(n=0)^oo(-1)^n (x^2)^(2n)/((2n)!)`                    

              `=sum_(n=0)^oo (-1)^n x^(4n)/((2n)!)`

              `= 1-x^4/2+x^8/24-x^(12)/720+ x^16/40320` -...

The integral becomes:

`int_0^1 cos(x^2) dx =int_0^1 [ 1-x^4/2+x^8/24-x^(12)/720+ x^16/40320-...]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^4/2+x^8/24-x^(12)/720+ x^16/40320-...]dx`

`=x-x^5/(2*5)+x^9/(24*9)-x^(13)/(720*13)+ x^17/(40320*17)-` ...

`=x-x^5/10+x^9/216-x^(13)/9360+ x^17/685440-` ...

Apply definite integral formula: `F(x)|_a^b = F(b) - F(a)` .

`F(1)=1-1^5/10+1^9/216-1^(13)/9360+ 1^17/685440-` ...

      `=1-1/10+1/216-1/9360+ 1/685440-` ...

`F(0) =0-0^5/10+0^9/216-0^(13)/9360+ 0^17/685440-` ...

      `= 0-0+0-0+0-` ...

All the terms are 0 then `F(0)= 0` .

We can stop at `5th` term  `(1/685440~~ 0.0000014589)` since we only need an error less than `0.0001` .

Then,

`F(1)-F(0)= [1-1/10+1/216-1/9360+ 1/685440-]-[0]`

           ` = 0.9045242509`

Thus, the approximated integral value is:

`int_0^1 cos(x^2) dx ~~0.9045`