`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.
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`
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