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`int_(0.1)^(0.3) sqrt(1+x^3) 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_0.1^0.3 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` .       

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