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Wize AP Calculus (AB) Textbook > Integration & Accumulation of Change

Initial Conditions

Antiderivatives of 1x\frac{1}{x}Previous Section
The Fundamental Theorem of Calculus - Part 1Next Section
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Initial Conditions

We don't always get unique answers for ∫f(x)dx\displaystyle \int f(x)dx. Instead we get F(x)+CF(x) +C. If we are given some initial conditions for the function, we can find this constant by substituting the coordinates and solving for CC .

Finding f(x) from f'(x)

To recover f(x)f(x) from f′(x)f'(x), givenf(a)=bf(a)=b:
  1. Antiderive f′(x)f'(x)to obtain f(x)f(x). (Don't forget +C+C)
  2. Substitute f(a)=bf(a)=b to solve for CC.
  3. Rewrite f(x)f(x) with the value ofCC.

Finding f(x) from f''(x)

To recover f(x)f(x) from f′′(x)f''(x), given two initial conditions, we have to integrate twice:
  1. Antiderive f′′(x)f''(x) to obtain f′(x)f'(x). (Don't forget +C+C)
  2. Antiderive f′(x)f'(x) to obtain f(x)f(x).
  3. The second time we antiderive we pick up another constant of integration DD , and we get Cx+DCx+D.
  4. Substitute the given conditions to find CC and DD.
  5. Rewrite f(x)f(x)with the values of CC and DD.
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Example: Initial Conditions

Find f(x)f(x) if f(0)=1f(0)=1 and f′(x)=x2+xf'(x)=x^2+x.

f(x)=∫(x2+x)dxf(x)=\displaystyle \int (x^2+x) dx

f(x)=x33+x22+Cf(x)\displaystyle =\frac{x^3}{3}+\frac{x^2}{2}+C

Now substitutte the intial condition f(0)=1f(0)=1

  ⟹  033+022+C=1\implies \displaystyle \frac{0^3}{3}+\frac{0^2}{2}+C=1

  ⟹  C=1\implies C=1

f(x)=x33+x22+1\displaystyle \boxed{f(x)=\frac{x^3}{3}+\frac{x^2}{2}+1}
Given f′′(x)=5ex+2sin⁡xf''(x)=5e^x+2\sin x, f(0)=0f(0)=0, and f(π)=0f(\pi)=0, find f(x)f(x).