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Question
If f'(x) = x2 + 5 and f(0) = −1, then find the value of f(x).
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Solution
f'(x) = x2 + 5 ...(Given)
∴ f(x) = ∫f'(x) dx
∴ f(x) = ∫(x2 + 5) dx
∴ f(x) = ∫ x2 dx + 5 ∫ dx
∴ f(x) = `"x"^3/3 + 5"x" + "c"` ....(i)
Substitute x = 0, f(0) = −1 ...(Given)
∴ f(x) = `"x"^3/3 + 5"x" + "c"`
∴ f(0) = `0^3/3 + 5(0) + "c"`
∴ −1 = 0 + 0 + c
∴ c = −1
Substituting c = – 1 in (i), we get,
∴ f(x) = `"x"^3/3 + 5"x" + (− 1)`
∴ f(x) = `"x"^3/3 + 5"x" − 1`
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