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f′′(x)=x−2+x3+2,f′(1)=14,f(1)=0.f(x)−?f''(x)=x^{-2}+x^3+2, f'(1)=\frac{1}{4}, f(1)=0. f(x)-?

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Q. f′′(x)=x−2+x3+2,f′(1)=14,f(1)=0.f(x)−?f''(x)=x^{-2}+x^3+2, f'(1)=\frac{1}{4}, f(1)=0. f(x)-?
  1. Integrate f′′(x)f''(x): Given f′′(x)=x−2+x3+2f''(x) = x^{-2} + x^3 + 2, we need to integrate to find f′(x)f'(x). Integrate f′′(x)f''(x) to get f′(x)f'(x): ∫(x−2+x3+2) dx=−x−1+14x4+2x+C\int(x^{-2} + x^3 + 2) \, dx = -x^{-1} + \frac{1}{4}x^4 + 2x + C, where CC is the constant of integration.
  2. Find CC: We know f′(1)=14f'(1) = \frac{1}{4}. Plug x=1x = 1 into f′(x)f'(x) to find CC.\newline−1+14+2+C=14-1 + \frac{1}{4} + 2 + C = \frac{1}{4}.
  3. Solve for C: Solve for C: C=14−2+1=−34+1=14C = \frac{1}{4} - 2 + 1 = -\frac{3}{4} + 1 = \frac{1}{4}.

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