# what is primitive F(x)=(px+q)(4-x^2)^(1/2)+r*arcsin(x/2) if f(x)=(4-x^2)^(1/2)?

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### 1 Answer

You need to use the relation between a function and its primitive, such that:

`F'(x) = f(x)`

Hence, you need to differentiate the function `F(x)` with respect to `x` , such that:

`F'(x) = (px+q)'(4-x^2)^(1/2) + (px+q)((4-x^2)^(1/2))' + r*(arcsin(x/2))'`

`F'(x) = p*sqrt(4 - x^2) + (px + q)*(1/2)(4 - x^2)^(1/2 - 1)*(4 - x^2)' + r*(1/2)/(sqrt(1 - x^2/4))`

`F'(x) = p*sqrt(4 - x^2) + (px + q)*(-2x)/(sqrt(4 - x^2)) + r/(2sqrt(1 - x^2/4))`

`F'(x) = p*sqrt(4 - x^2) + (px + q)*(-2x)/(sqrt(4 - x^2)) + r/(2(sqrt(4 - x^2))/2)`

`F'(x) = p*sqrt(4 - x^2) + (px + q)*(-2x)/(sqrt(4 - x^2)) + r/(sqrt(4 - x^2))`

`F'(x) = (p*(4 - x^2) - 2px^2 - 2qx + r)/(sqrt(4 - x^2))`

`F'(x) = (4p - px^2 - 2px^2 - 2qx + r)/(sqrt(4 - x^2))`

`F'(x) = (x^2(-p - 2p) - 2qx + r + 4p)/(sqrt(4 - x^2))`

`F'(x) = ((x^2(-p - 2p) - 2qx + r + 4p)(sqrt(4 - x^2)))/(4 - x^2)`

`F'(x) = ((-3px^2 - 2qx + r + 4p)(sqrt(4 - x^2)))/(4 - x^2)`

`F'(x) = ((-3px^2 - 2qx + r + 4p)/(4 - x^2))*sqrt(4 - x^2)`

You need to set equal the equations of `F'(x)` and `f(x)` , such that:

`((-3px^2 - 2qx + r + 4p)/(4 - x^2))*sqrt(4 - x^2) = sqrt(4 - x^2)`

Equating the coefficients of `sqrt(4 - x^2)` both sides, yields:

`((-3px^2 - 2qx + r + 4p)/(4 - x^2)) = 1`

`-3px^2 - 2qx + r + 4p = 4 - x^2`

Equating the coefficients of like powers yields:

`-3p = -1 => 3p = 1 => p = 1/3`

`-2q = 0 => q = 0`

`r + 4p = 4 => r + 4/3 = 4 => r = 4 - 4/3 => r = 8/3`

**Hence, evaluating the primitive `F(x) = (px + q)sqrt(4 - x^2) + r*arcsin(x/2),` under the given conditions, yields **`F(x) = (x/3)sqrt(4 - x^2) + (8/3)*arcsin(x/2).`