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Question
If f'(x) = 4x3 − 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
Solution
f'(x) = 4x3 − 3x2 + 2x + k ....[Given]
f(x) = ∫ f'(x) dx
= ∫ (4x3 − 3x2 + 2x + k) dx
= 4 ∫ x3 dx − 3 ∫ x2 dx + 2 ∫ x dx + k ∫ dx
= `4 ("x"^4/4) - 3("x"^3/3) + 2("x"^2/2) "kx" + "c"`
∴ f(x) = x4 − x3 + x2 + kx + c ....(i)
Now, f(0) = 1 ...[Given]
∴ (0)4 − (0)3 + (0)2 + k(0) + c = 1
∴ c = 1 ....(ii)
Also, f(1) = 4 ....[Given]
∴ 1 − 1 + 1 + k + c = 4
∴ 1 + k + 1 = 4
∴ 2 + k = 4
∴ k = 2 ...(iii)
Substituting (ii) and (iii) in (i), we get
f(x) = x4 − x3 + x2 + 2x + 1
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