Calculus of a Single Variable, Chapter 9, 9.10, Section 9.10, Problem 70

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