`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.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial