AP Calculus AB · Lesson 3

Antiderivatives and substitution

Know Know it

An antiderivative of f is a function F with F′=fF' = f. The indefinite integral ∫f(x) dx=F(x)+C\int f(x)\,dx = F(x) + C names all of them at once.

Basic rules. ∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C (n≠−1n \neq -1); ∫1x dx=ln⁡∣x∣+C\int \frac{1}{x}\,dx = \ln|x| + C; ∫ex dx=ex+C\int e^x\,dx = e^x + C; ∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + C; ∫cos⁡x dx=sin⁡x+C\int \cos x\,dx = \sin x + C; ∫sec⁡2x dx=tan⁡x+C\int \sec^2 x\,dx = \tan x + C. Constants factor out, and sums integrate term by term.

Initial condition. If you know one value, such as f(2)=5f(2) = 5, put it into F(x)+CF(x) + C to solve for C.

Substitution reverses the chain rule. Choose uu as an inner function whose derivative (up to a constant factor) also appears. Replace it with du=u′(x) dxdu = u'(x)\,dx, integrate in uu, then switch back to xx. In a definite integral, either change the limits to uu-values or switch back to xx before evaluating.

Trap Keep the limits matched to the variable

After substituting u=x2+1u = x^2 + 1 in ∫02… dx\int_0^2 \dots\,dx, the limits become u=1u = 1 and u=5u = 5. Evaluating the uu-antiderivative from 0 to 2 mixes up the variables and gives a wrong answer. Also keep any constant factor: if du=2x dxdu = 2x\,dx, then x dx=12dux\,dx = \frac{1}{2}du.

Worked example Easy

A function ff has f′(x)=4x−3f'(x) = 4x - 3 and f(2)=5f(2) = 5. What is f(5)f(5)?

  1. 38Answer
  2. 17
  3. 35
  4. 40

How to solve it

  1. Antiderivative: f(x)=2x2−3x+Cf(x) = 2x^2 - 3x + C.
  2. Use f(2)=5f(2) = 5: 8−6+C=58 - 6 + C = 5, so C=3C = 3.
  3. f(5)=2(25)−15+3=38f(5) = 2(25) - 15 + 3 = 38.

Why each choice is right or wrong

  • A. Correct. f(x)=2x2−3x+Cf(x) = 2x^2 - 3x + C, and f(2)=2+C=5f(2) = 2 + C = 5 gives C=3C = 3, so f(5)=50−15+3=38f(5) = 50 - 15 + 3 = 38.
  • B. Incorrect. 17 is f′(5)=20−3f'(5) = 20 - 3, the slope at x=5x = 5, not the value of ff.
  • C. Incorrect. 35 is 2(5)2−3(5)2(5)^2 - 3(5), which leaves out the constant C=3C = 3 found from f(2)=5f(2) = 5.
  • D. Incorrect. 40 uses C=5C = 5, the value of f(2)f(2), without solving 2+C=52 + C = 5 for the constant.

Explain Explain it

Task verbs: calculate, explain

Show the substitution in writing: name uu, write dudu, rewrite the integral entirely in uu (with new limits if it is definite), then integrate. A reader should see every step.

Check an antiderivative by differentiating it; you should get the integrand back.

Worked example Medium

What is ∫016x(x2+1)2 dx\int_0^1 6x(x^2 + 1)^2\,dx?

  1. 1
  2. 73\frac{7}{3}
  3. 7Answer
  4. 14

How to solve it

  1. Let u=x2+1u = x^2 + 1, so du=2x dxdu = 2x\,dx and 6x dx=3 du6x\,dx = 3\,du.
  2. New limits: x=0x = 0 gives u=1u = 1; x=1x = 1 gives u=2u = 2.
  3. 3∫12u2 du=[u3]12=8−1=73\int_1^2 u^2\,du = [u^3]_1^2 = 8 - 1 = 7.

Why each choice is right or wrong

  • A. Incorrect. 1 keeps the x-limits 0 and 1 for u, computing 3∫01u2 du3\int_0^1 u^2\,du; the u-limits are 02+1=10^2 + 1 = 1 and 12+1=21^2 + 1 = 2.
  • B. Incorrect. 73\frac{7}{3} is ∫12u2 du\int_1^2 u^2\,du, which drops the factor 3 that comes from 6x dx=3 du6x\,dx = 3\,du.
  • C. Correct. With u=x2+1u = x^2 + 1, du=2x dxdu = 2x\,dx, so 6x dx=3 du6x\,dx = 3\,du and the integral is 3∫12u2 du=[u3]12=8−1=73\int_1^2 u^2\,du = [u^3]_1^2 = 8 - 1 = 7.
  • D. Incorrect. 14 is 6∫12u2 du6\int_1^2 u^2\,du, which replaces 6x dx6x\,dx with 6 du6\,du and forgets that du=2x dxdu = 2x\,dx.

Calculate Your turn

Use substitution to find ∫02xex2 dx\int_0^2 x e^{x^2}\,dx. Show your choice of uu, how the limits change, and the value.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Chooses u=x2u = x^2 with du=2x dxdu = 2x\,dx, so x dx=12dux\,dx = \frac{1}{2}du
  • Changes the limits to u=0u = 0 and u=4u = 4, or switches back to xx before evaluating
  • Finds the value e4−12\frac{e^4 - 1}{2}, about 26.8

Model answer

Let u=x2u = x^2, so du=2x dxdu = 2x\,dx and x dx=12dux\,dx = \frac{1}{2}du. When x=0x = 0, u=0u = 0; when x=2x = 2, u=4u = 4. Then ∫02xex2 dx=12∫04eu du=12(e4−1)≈26.80\int_0^2 x e^{x^2}\,dx = \frac{1}{2}\int_0^4 e^u\,du = \frac{1}{2}(e^4 - 1) \approx 26.80.

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