Calculus of a Single Variable, Chapter 9, 9.10, Section 9.10, Problem 69

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_0.1^0.3 sqrt(1+x^3)dx , we let:
sqrt(1+x^3) =(1+x^3)^(1/2)
Using the aforementioned power series, we may replace the "x " with "x^3 " and "k " with "1/2 or 0.5" .
(1+x^3)^(1/2) =1 +0.5x^3+ (0.5(0.5-1))/2! (x^3)^2 + ...
=1 +0.5x^3 -0.25/2! x^6 + ...
The integral becomes:
int_0.1^0.3 sqrt(1+x^3)dx=int_0.1^0.3 [1 +0.5x^3-0.25/2! x^6 +...]dx
To determine the indefinite integral, we integrate each term using Power Rule for integration: int x^ndx =x^(n+1)/(n+1).
int_0.1^0.3 [1 +0.5x^3-0.25/2! x^6 +...]dx
=[x + 0.5 x^4/4-0.25/2! x^7/7 +...]|_0.1^0.3
=[ x +x^4/8-x^7/56]_0.1^0.3
Apply definite integral formula: F(x)|_a^b = F(b) - F(a) .
F(0.3) =0.3+0.3^4/8-0.3^7/56+ ...
=0.3 +0.0010125 -3.9054x10^(-6) + ...
F(0.1) = 0.1 +0.1^4/8-0.1^7/56+ ...
= 0.1+.0000125 - 1.7857x10^(-9) + ...
We stop at the 3rd term since we only need error less than 0.0001 .
F(0.3)- F(0.1) = [0.3 +0.0010125-3.9054x10^(-6)] -[0.1+.0000125-1.7857x10^(-9) ]
=0.2009960964
Thus, int_0.1^0.3 sqrt(1+x^3)dx ~~0.2010 .