calculus II

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The interval of convergence of the power series `Sigma_(k=1)^n x^n` is `|x| < 1 => -1 < x < 1` and the radius of convergence of power series is `R = 1` .

You need to evaluate the limit of the series `lim_(n->oo) (a_n)` , hence, you need to evaluate the definite integral `int_0^1f(x) dx,` where` f(x) = 3x/(2 + x^3)` .

You need to use partial fraction decomposition to evaluate the definite integral, such that:

`int_0^1 3x/(2 + x^3) dx = int_0^1 3x/(x + root(3)2)(x^2 - x*root(3)2 + root(3)4) dx`

`3x/(x + root(3)2)(x^2 - x*root(3)2 + root(3)4) = a/(x + root(3)2) + (bx + c)/(x^2 - x*root(3)2 + root(3)4)`

`3x = x^2(a+b) + x(-a*root(3)2 + b*root(3)2 + c) + a*root(3)4 + c*root(3)2`

Equating the coefficients of like parts, yields:

`{(a+b = 0),(-a*root(3)2 + b*root(3)2 + c = 3),(a*root(3)4 + c*root(3)2 = 0):}`

`{(-a = b),(2b*root(3)2+ c = 3),(-b*root(3)2 + c = 0):} => {(-a = b),(3b*root(3)2= 3),(-b*root(3)2 + c = 0):}`

`{(a = -1/(root(3)2),(b = 1/(root(3)2)),(c = 1):}`

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

Integrating both sides, yields:

`int_0^1 3x/(2 + x^3) dx = (1/(root(3)2)(int_0^1 -1/(x + root(3)2) dx + int_0^1 (x + root(3)2)/(x^2 - x*root(3)2 + root(3)4)dx)`

`int_0^1 3x/(2 + x^3) dx = (1/(root(3)2)(ln(root(3)2) - ln(1 + root(3)2) + int_0^1 (x + root(3)2)/(x^2 - x*root(3)2 + root(3)4)dx)`

`int_0^1 (x + root(3)2)/(x^2 - x*root(3)2 + root(3)4)dx) = (1/2)int_0^1 (2x + 2root(3)2)/(x^2 - x*root(3)2 + root(3)4)dx)`

`(1/2)int_0^1 (2x - root(3)2)/(x^2 - x*root(3)2 + root(3)4)dx + (3root(3)2)/2 int_0^1 1/(x^2 - x*root(3)2 + root(3)4)dx`

`int_0^1 (x + root(3)2)/(x^2 - x*root(3)2 + root(3)4)dx) = (1/2) (ln(1 - root(3)2 + root(3)4) - ln(root(3)4)) +(3root(3)2)/2 int_0^1 1/((x - root(3)2)^2 + (3root(3)4)/4)dx `

`int_0^1 (x + root(3)2)/(x^2 - x*root(3)2 + root(3)4)dx) = (1/2) (ln(1 - root(3)2 + root(3)4) - ln(root(3)4)) +(3root(3)2)/2 *2/(sqrt3*root(3)2) (tan^(-1)(1 - root(3)2)/(sqrt3*root(3)2 + tan^(-1) (root(3)2)/(sqrt3*root(3)2)`

`int_0^1 3x/(2 + x^3) dx = (1/(root(3)2)(ln(root(3)2) - ln(1 + root(3)2) + (1/2) (ln(1 - root(3)2 + root(3)4) - ln(root(3)4)) +(3root(3)2)/2 *2/(sqrt3*root(3)2) (tan^(-1)(1 - root(3)2)/(sqrt3*root(3)2 + tan^(-1) (root(3)2)/(sqrt3*root(3)2)`

`lim_(n->oo)(a_n) = int_0^1 3x/(2 + x^3) dx = (1/(root(3)2)(ln(root(3)2) - ln(1 + root(3)2) + (1/2) (ln(1 - root(3)2 + root(3)4) - ln(root(3)4)) +(3root(3)2)/2 *2/(sqrt3*root(3)2) (tan^(-1)(1 - root(3)2)/(sqrt3*root(3)2 + tan^(-1) (root(3)2)/(sqrt3*root(3)2)`

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