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Calculus Honors · AP-aligned study guide

4.1 · Antiderivatives and initial conditions

Reverse differentiation, retain the family of constants, and choose a particular antiderivative from a given condition.

AP 6.8; 7.7 · Reading: Larson & Edwards §§4.1, 5.2, 5.4

Learning objectives

  • Find an antiderivative and verify it by differentiation.
  • Use initial data to determine a constant.
  • Separate the power-rule case from the logarithmic case.

An indefinite integral describes a family

If F′(x)=f(x)\displaystyle F'(x)=f(x), then F\displaystyle F is an antiderivative of f\displaystyle f. Adding any constant leaves the derivative unchanged, so an indefinite integral includes +C\displaystyle +C.

For n≠−1\displaystyle n\ne-1, increase the power by one and divide by that new power. Verify the result by differentiating rather than relying on its appearance.

∫xn dx=xn+1n+1+C,n≠−1\int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\qquad n\ne-1

Recognize the logarithmic exception

The power rule cannot divide by zero when the integrand is x−1\displaystyle x^{-1}. Instead, ∫(1/x) dx=ln⁡∣x∣+C\displaystyle \int (1/x)\,dx=\ln|x|+C on intervals excluding zero. The absolute value allows either positive or negative input intervals.

Other basic pairs include ∫ex dx=ex+C\displaystyle \int e^x\,dx=e^x+C, ∫cos⁡x dx=sin⁡x+C\displaystyle \int\cos x\,dx=\sin x+C, and ∫sin⁡x dx=−cos⁡x+C\displaystyle \int\sin x\,dx=-\cos x+C. Constants multiply the whole antiderivative.

Fit the initial condition after integrating

If y′=2x\displaystyle y'=2x and y(1)=4\displaystyle y(1)=4, integration gives y=x2+C\displaystyle y=x^2+C. Substituting the point gives 4=1+C\displaystyle 4=1+C, hence C=3\displaystyle C=3. The initial condition selects one member of the family.

If acceleration is given and position is requested, integrate twice, with a different constant at each stage. Velocity data determines the first constant; position data determines the second.

Worked examples

Integrate a sum

Find ∫(3x2+2) dx\displaystyle \int(3x^2+2)\,dx.

  1. Integrate term by term.
  2. An antiderivative is x3+2x+C\displaystyle x^3+2x+C.
  3. Differentiating gives 3x2+2\displaystyle 3x^2+2, confirming the result.

Fit a condition

Solve y′=3x2\displaystyle y'=3x^2 with y(1)=5\displaystyle y(1)=5.

  1. Integration gives y=x3+C\displaystyle y=x^3+C.
  2. The condition implies 5=1+C\displaystyle 5=1+C.
  3. Thus y=x3+4\displaystyle y=x^3+4.

Common mistakes

  • Forgetting the integration constant.
  • Applying the power antiderivative rule to 1/x.
  • Using only one constant when integrating twice.

Try it yourself

No calculator required.

1. Find ∫(4x3−2x+1) dx\displaystyle \int(4x^3-2x+1)\,dx and check it.

Show worked solution

The family is x4−x2+x+C\displaystyle x^4-x^2+x+C. Its derivative is 4x3−2x+1\displaystyle 4x^3-2x+1.

2. Find ∫(2/x+ex) dx\displaystyle \int(2/x+e^x)\,dx on an interval excluding zero.

Show worked solution

The result is 2ln⁡∣x∣+ex+C\displaystyle 2\ln|x|+e^x+C. Differentiation gives 2/x+ex\displaystyle 2/x+e^x on either permitted side of zero.

3. Solve s′′(t)=6t\displaystyle s''(t)=6t with s′(0)=2\displaystyle s'(0)=2 and s(0)=1\displaystyle s(0)=1.

Show worked solution

Integrate once: s′(t)=3t2+C1\displaystyle s'(t)=3t^2+C_1, so C1=2\displaystyle C_1=2. Integrate again: s(t)=t3+2t+C2\displaystyle s(t)=t^3+2t+C_2, so C2=1\displaystyle C_2=1. The result is t3+2t+1\displaystyle t^3+2t+1.

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