`int_0^(0.2) sqrt(1+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:

`(1+x)^k = 1 +kx+ (k(k-1))/2! x^2 +(k(k-1)(k-2))/3!x^3 +` ...

 To apply this on the given integral `int_0^0.2 sqrt(1+x^2)dx` , we let:

`sqrt(1+x^2) =(1+x^2)^(1/2)`

 Using the aforementioned power series, we may replace the "`x` " with "`x^2` " and "`k` " with "`1/2 or 0.5` ".

`(1+x^2)^(1/2) =1 +0.5x^2+ (0.5(0.5-1))/2! (x^2)^2 +...`

             ` = 1 +0.5x^2 -0.25/2! x^4 +...`

             `= 1 +x^2/2-x^4/8 +...`

The integral becomes:

`int_0^0.2 sqrt(1+x^2)dx = int_0^0.2[1 +x^2/2-x^4/8 +...]dx`

To determine the indefinite integral, we integrate each term using Power Rule for integration: `int x^ndx =x^(n+1)/(n+1)` .

`int_0^0.2[1 +x^2/2-x^4/8 +...]dx = [x +x^3/(2*3) -x^5/(8*5) +...]|_0^0.2`

                `= [x +x^3/6 -x^5/40+...]|_0^0.2`

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

`F(0.2)=0.2 +0.2^3/6 -0.2^5/40+` ...

             `=0.2+1.3333x10^(-3)-8x10^(-6)+` ...

`F(0) =0+0^3/6-0^5/40+` ...

          `= 0+0-0+...`

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

We can stop at 3rd term `(8x10^(-6) or 0.000008)` since we only need an error less than `0.0001` .

Then,

`F(0.2)-F(0) = [0.2+1.3333x10^(-3)-8x10^(-6)] -[0]`

                         `= 0.2013253`

Thus, the approximated integral value is:

`int_0^0.2 sqrt(1+x^2)dx ~~0.2013`