Question

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

Answer 1

f'(x) = x² + 5         ...(Given)

∴ f(x) = ∫f'(x) dx

∴ f(x) = ∫(x² + 5) dx

∴ f(x) = ∫ x² dx + 5 ∫ dx

∴ f(x) = ${\displaystyle \frac{{\text{x}}^3}{3}+5\text{x}+\text{c}}$     ....(i)

Substitute x = 0, f(0) = −1         ...(Given)

∴ f(x) = ${\displaystyle \frac{{\text{x}}^3}{3}+5\text{x}+\text{c}}$

∴ f(0) = ${\displaystyle \frac{0^3}{3}+5\left(0\right)+\text{c}}$

∴ −1 = 0 + 0 + c

∴ c = −1

Substituting c = – 1 in (i), we get,

∴ f(x) = ${\displaystyle \frac{{\text{x}}^3}{3}+5\text{x}+(-1)}$

∴ f(x) = ${\displaystyle \frac{{\text{x}}^3}{3}+5\text{x}-1}$