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...

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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` .