`int_0^(1/4) xln(x+1) dx` Use a power series to approximate the value of the integral with an error of less than 0.0001.

Expert Answers

From the basic list of power series, we have:

`ln(x) =sum_(n=0)^oo (-1)^(n) (x-1)^(n+1)/(n+1)`

`= (x-1)-(x-1)^2/2+(x-1)^3/3 -(x-1)^4/4 +...`

We replace "`x` " with "`x+1` " to setup:

`ln(1+x) =sum_(n=0)^oo (-1)^n ((x+1)-1)^(n+1)/(n+1)`

`=sum_(n=0)^oo (-1)^n x^(n+1)/(n+1)`

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From the basic list of power series, we have:

`ln(x) =sum_(n=0)^oo (-1)^(n) (x-1)^(n+1)/(n+1)`

`= (x-1)-(x-1)^2/2+(x-1)^3/3 -(x-1)^4/4 +...`

We replace "`x` " with "`x+1` " to setup:

`ln(1+x) =sum_(n=0)^oo (-1)^n ((x+1)-1)^(n+1)/(n+1)`

`=sum_(n=0)^oo (-1)^n x^(n+1)/(n+1)`

`=x-x^2/2+x^3/3 -x^4/4+...`

Note: `((x+1)-1) = (x+1-1) = x`

Then,

`x ln(1+x) =sum_(n=0)^oo (-1)^n x^(n+1)/(n+1) *x`

`=sum_(n=0)^oo (-1)^n x^(n+2)/(n+1)`

Note: `x^(n+1) * x = x^(n+1+1) =x^(n+2)`

Applying the summation formula, we get:

`x ln(1+x)= x*[x-x^2/2+x^3/3 -x^4/4+...]`

or

`= x^2 -x^3/2+x^4/3-x^5/4 +...`

Then the integral becomes:

`int_0^(1/4) xln(x+1) = int_0^(1/4) [x^2 -x^3/2+x^4/3-x^5/4 +...]dx`

To determine the indefinite integral, we integrate each term using the Power Rule for integration: `int x^n dx= x^(n+1)/(n+1)` .

`int_0^(1/4) [x^2 -x^3/2+x^4/3-x^5/4 +...]dx`

` = [x^3/3 -x^4/(2*4)+x^5/(3*5)-x^6/(4*6) +...]_0^(1/4)`

` = [x^3/3 -x^4/7+x^5/15-x^6/24 +...]_0^(1/4)`

Apply definite integral formula: `F(x)|_a^b = F(b) - F(a)` .

`F(1/4) or F(0.25) =0.25^3/3 -0.25^4/7+0.25^5/15-0.25^6/24 +...`

`=1/192-1/1792+1/15360 -1/98304+...`

`F(0)=0^3/3 -0^4/7+0^5/15-0^6/24 +...`

`= 0-0+0-0+...`

All the terms are 0 then `F(0) =0` .

We may stop at `4th` term `(1/98304~~0.00001017)` since we only need an error less than `0.0001` .

`F(1/4)-F(0) = [1/192-1/1792+1/15360 -1/98304]-[0]`

`= 0.00470522926`

Thus, the approximated integral value:

`int_0^(1/4) xln(x+1) dx ~~0.0047`

Approved by eNotes Editorial Team