AP Calculus AB · Lesson 2

The Fundamental Theorem and accumulation

Know Know it

An accumulation function has a variable upper limit: g(x)=∫axf(t) dtg(x) = \int_a^x f(t)\,dt. It adds the signed area under f from a to x.

Fundamental Theorem, derivative form. If f is continuous, then g′(x)=f(x)g'(x) = f(x). With a function as the upper limit, use the chain rule: ddx∫au(x)f(t) dt=f(u(x)) u′(x)\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x))\,u'(x).

Fundamental Theorem, evaluation form. If f is continuous on [a,b][a, b] and F is any antiderivative of f, then ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a).

Net change. Applied to a rate, the theorem says ∫abQ′(t) dt=Q(b)−Q(a)\int_a^b Q'(t)\,dt = Q(b) - Q(a), so the amount at time b is Q(b)=Q(a)+∫abQ′(t) dtQ(b) = Q(a) + \int_a^b Q'(t)\,dt.

Properties. ∫aaf=0\int_a^a f = 0; ∫baf=−∫abf\int_b^a f = -\int_a^b f; ∫abf+∫bcf=∫acf\int_a^b f + \int_b^c f = \int_a^c f; ∫abkf=k∫abf\int_a^b k f = k\int_a^b f.

Trap An accumulation function is not the integrand

For g(x)=∫0xf(t) dtg(x) = \int_0^x f(t)\,dt, the value g(4)g(4) is an area, but g′(4)=f(4)g'(4) = f(4) is a height. Read a graph of f for heights when you need g′g', and add areas when you need gg. Areas below the axis count as negative.

Worked example Easy

What is ∫13(4x+1) dx\int_1^3 (4x + 1)\,dx?

  1. 8
  2. 16
  3. 21
  4. 18Answer

How to solve it

  1. Antiderivative: F(x)=2x2+xF(x) = 2x^2 + x.
  2. F(3)=18+3=21F(3) = 18 + 3 = 21 and F(1)=2+1=3F(1) = 2 + 1 = 3.
  3. ∫13(4x+1) dx=21−3=18\int_1^3 (4x + 1)\,dx = 21 - 3 = 18.

Why each choice is right or wrong

  • A. Incorrect. 8 is f(3)−f(1)=13−5f(3) - f(1) = 13 - 5, which subtracts values of the integrand instead of an antiderivative.
  • B. Incorrect. 16 uses only the 2x22x^2 part of the antiderivative, 18−218 - 2, and drops the xx from integrating the 1.
  • C. Incorrect. 21 is F(3)F(3) alone; the theorem subtracts F(1)=3F(1) = 3.
  • D. Correct. An antiderivative is F(x)=2x2+xF(x) = 2x^2 + x, so F(3)−F(1)=21−3=18F(3) - F(1) = 21 - 3 = 18.

Explain Explain it

Task verbs: interpret, calculate

Interpret an integral of a rate as a change in the amount, with units and the time span: "∫05R(t) dt=40\int_0^5 R(t)\,dt = 40 means 40 gallons flowed in during the first 5 hours."

Calculate an amount as the starting amount plus the net change. Do not forget the starting value.

Worked example Medium

Let g(x)=∫0x21+2t dtg(x) = \int_0^{x^2} \sqrt{1 + 2t}\,dt. What is g′(2)g'(2)?

  1. 3
  2. 12Answer
  3. 454\sqrt{5}
  4. 263\frac{26}{3}

How to solve it

  1. The upper limit is u(x)=x2u(x) = x^2, so g′(x)=1+2u⋅u′(x)=1+2x2⋅2xg'(x) = \sqrt{1 + 2u}\cdot u'(x) = \sqrt{1 + 2x^2}\cdot 2x.
  2. At x=2x = 2: 1+8=3\sqrt{1 + 8} = 3 and 2x=42x = 4.
  3. g′(2)=3⋅4=12g'(2) = 3 \cdot 4 = 12.

Why each choice is right or wrong

  • A. Incorrect. 3 is 1+2(4)\sqrt{1 + 2(4)}, the integrand at the upper limit, without the chain-rule factor 2x=42x = 4.
  • B. Correct. g′(x)=1+2x2⋅2xg'(x) = \sqrt{1 + 2x^2} \cdot 2x, so g′(2)=9⋅4=12g'(2) = \sqrt{9} \cdot 4 = 12.
  • C. Incorrect. 454\sqrt{5} puts x=2x = 2 into the integrand instead of the upper limit x2=4x^2 = 4.
  • D. Incorrect. 263\frac{26}{3} is g(2)=∫041+2t dtg(2) = \int_0^4 \sqrt{1 + 2t}\,dt, the value of the function, not its derivative.

Interpret Your turn

Sand is added to a pile at a rate of S(t)S(t) cubic meters per hour, where tt is hours after noon, and no sand is removed. At noon the pile holds 50 cubic meters, and ∫04S(t) dt=36\int_0^4 S(t)\,dt = 36. Say what ∫04S(t) dt=36\int_0^4 S(t)\,dt = 36 means, and find how much sand is in the pile at 4 p.m.

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A full answer does this

  • Says 36 cubic meters of sand are added to the pile from noon to 4 p.m.
  • Finds that the pile holds 86 cubic meters at 4 p.m.

Model answer

∫04S(t) dt=36\int_0^4 S(t)\,dt = 36 means that 36 cubic meters of sand are added to the pile between noon and 4 p.m. Since the pile starts with 50 cubic meters and nothing is removed, at 4 p.m. it holds 50+36=8650 + 36 = 86 cubic meters of sand.

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