If f '(x) = x22-kx+1, f(0) = 2 and f(3) = 5, find f(x). - Mathematics and Statistics

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Sum

If f '(x) = `"x"^2/2 - "kx" + 1`, f(0) = 2 and f(3) = 5, find f(x).

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Solution

f '(x) = `"x"^2/2 - "kx" + 1`      ...[Given]

f(x) = ∫ f '(x) dx

`= int ("x"^2/2 - "kx" + 1)`dx

`= 1/2 int "x"^2 "dx" - "k" int "x" "dx" + int 1 * "dx"`

`= 1/2 * "x"^3/3 - "k" ("x"^2/2) + "x" + "c"`

∴ f(x) = `"x"^3/6 - "k"/2  "x"^2 + "x" + "c"`    ...(i)

Now, f(0) = 2

∴ `(0)^3/6 - "k"/2 (0)^2 + 0 + "c"` = 2

∴ c = 2       ...(ii)

Also f(3) = 5       ...[Given]

∴ `(3)^3/6 - "k"/2 (3)^2 + 3 + 2 = 5`

∴ `27/6 - "9k"/2 + 5 = 5`

∴ `9/2 - "9k"/2 = 0` 

∴ `"9k"/2 = 9/2`

∴ k = 1        ....(iii)

Substituting (ii) and (iii) in (i), we get

f(x) = `"x"^3/6 - "x"^2/2 + "x" + 2`

  Is there an error in this question or solution?
Chapter 5: Integration - EXERCISE 5.1 [Page 119]

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