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If f'(x) = x2 + 5 and f(0) = −1, then find the value of f(x).

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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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Chapter 5: Integration - EXERCISE 5.1 [Page 119]

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