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Question
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`
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