calculus II

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The function `f(x) = 3/(2 + x^3)` models the Riemann summation `a_n = 1/n*Sigma_(k=1)^n f(k/n)` such that:

`lim_(n->oo)(a_n) = int_0^1 f(x)dx`

`1/n*Sigma_(k=1)^n f(k/n) = 1/n*Sigma_(k=1)^n 3/(2 + (k/n)^3)`

You need to evaluate the limit of the sequence (a_n), hence, you need to evaluate the definite integral `int_0^1 3/(2 + x^3)...

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The function `f(x) = 3/(2 + x^3)` models the Riemann summation `a_n = 1/n*Sigma_(k=1)^n f(k/n)` such that:

`lim_(n->oo)(a_n) = int_0^1 f(x)dx`

`1/n*Sigma_(k=1)^n f(k/n) = 1/n*Sigma_(k=1)^n 3/(2 + (k/n)^3)`

You need to evaluate the limit of the sequence (a_n), hence, you need to evaluate the definite integral `int_0^1 3/(2 + x^3) dx` , using the partial fraction decomposition, such that:

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

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

`3 = a*root(3)4 - ax*root(3)2 + ax^2 + bx*root(3)2 + bx^2 + c*root(3)2 + cx`

`3 = 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 = -b`

`- a*root(3)2 + b*root(3)2 + c = 0 => 2 b*root(3)2 = -c`

`a*root(3)4 + c*root(3)2 = 3 => -b*root(3)4 -2broot(3)4 = 3 => -3broot(3)4 = 3 => b = -1/(root(3)4) => a = 1/(root(3)4) => c = 2/(root(3)2)`

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

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

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

`int_0^1 3/(2 + x^3) dx = (1/(root(3)4))(ln (root(3)2 + x)|_0^1 + int_0^1 (-x+ 2(root(3)2))/(root(3)4 - x*root(3)2 + x^2)))`

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

`root(3)4 - x*root(3)2 + x^2 = t => (2x - root(3)2)dx = dt`

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

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

`int_0^1 (-x+ 2(root(3)2))/(root(3)4 - x*root(3)2 + x^2))dx = (-1/2)ln (root(3)4 - x*root(3)2 + x^2))|_0^1 + (3(root(3)2))/2 int_0^1 1/((x - (root(3) 2))/2)^2 + sqrt3/2*root(3) 2) dx`

`int_0^1 (-x+ 2(root(3)2)/(root(3)4 - x*root(3)2 + x^2))dx = (-1/2)ln (root(3)4 - x*root(3)2 + x^2))|_0^1 + (3(root(3)2))/2*2/(sqrt3*(root(3) 2)) tan^(-1) (x - (root(3) 2))/2)/( sqrt3/2*root(3) 2)|_0^1`

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

Hence, evaluating the limit of the power series (a_n), using the evaluation of the integral yields `int_0^1 3/(2 + x^3) dx = (1/(root(3)4))(ln (root(3)2 + 1) - ln (root(3)2) + (-1/2)(ln (root(3)4 - root(3)2 + 1) - ln (root(3)4)) + (3(root(3)2))/2*2/(sqrt3*(root(3) 2)) (tan^(-1) (1- (root(3) 2))/2)/( sqrt3/2*root(3) 2) + tan^(-1) ((root(3) 2))/2)/( sqrt3/2*root(3) 2).`

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