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Integration

4.4 The Fundamental Theorem of Calculus

Focus on ftc part 1: accumulation functions, ftc part 2: endpoint evaluation, average value of a function, mean value theorem for integrals, net change versus total change.

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Ideas in this lesson

01FTC Part 1: accumulation functionsWorked steps · five questions02FTC Part 2: endpoint evaluationWorked steps · five questions03Average value of a functionWorked steps · five questions04Mean Value Theorem for IntegralsWorked steps · five questions05Net change versus total changeWorked steps · five questions
Idea 1 of 5

FTC Part 1: accumulation functions

Differentiating an integral with a variable upper endpoint returns the integrand there, times the endpoint derivative.

Worked example 1

Let F(x)=∫0xt2+2 dtF(x) = \int_{0}^{x} t^{2} + 2\,dt. Find F′(1)F'(1).
  1. F′(x)=x2+2F'(x) = x^{2} + 2
  2. F′(1)=3F'(1) = 3

Answer: 33

Worked example 2

Let F(x)=∫0x2t2+3 dtF(x) = \int_{0}^{x^{2}} t^{2} + 3\,dt. Find F′(2)F'(2).
  1. F′(x)=(x4+3)⋅2xF'(x) = (x^{4} + 3)\cdot 2x
  2. F′(2)=76F'(2) = 76

Answer: 7676

Idea 2 of 5

FTC Part 2: endpoint evaluation

If F′=fF' = f, then ∫\int ₐᵇ f=F(b)−F(a)f = F(b) - F(a) under the FTCFTC hypotheses.

Worked example 1

Evaluate ∫02x dx\int_{0}^{2} x\,dx.
  1. F(x)=x22F(x) = \dfrac{x^{2}}{2}
  2. F(2)−F(0)=222−0=2F(2) - F(0) = \dfrac{2^{2}}{2} - 0 = 2

Answer: 22

Worked example 2

Evaluate ∫03x2 dx\int_{0}^{3} x^{2}\,dx.
  1. F(x)=x33F(x) = \dfrac{x^{3}}{3}
  2. F(3)−F(0)=333−0=9F(3) - F(0) = \dfrac{3^{3}}{3} - 0 = 9

Answer: 99

Idea 3 of 5

Average value of a function

favg=(1b−a)∫f_{\mathrm{avg}} = (\frac{1}{b - a})\int ₐᵇ f(x)dxf(x) dx. Divide signed area by interval length.

Worked example 1

Find the average value of f(x)=xf(x) = x on [0,2][0, 2].
  1. favg=12∫02x dxf_{\mathrm{avg}} = \dfrac{1}{2} \int_{0}^{2} x\,dx
  2. ∫02xdx=222\int _{0}^{2} x dx = \dfrac{2^{2}}{2}
  3. favg=22f_{\mathrm{avg}} = \dfrac{2}{2}

Answer: 11

Worked example 2

Find the average value of f(x)=x2f(x) = x^{2} on [0,3][0, 3].
  1. favg=13∫03x2 dxf_{\mathrm{avg}} = \dfrac{1}{3} \int_{0}^{3} x^{2}\,dx
  2. ∫03x2dx=333\int _{0}^{3} x^{2} dx = \dfrac{3^{3}}{3}
  3. favg=323f_{\mathrm{avg}} = \dfrac{3^{2}}{3}

Answer: 33

Idea 4 of 5

Mean Value Theorem for Integrals

For continuous ff, some cc in [a,b][a, b] satisfies f(c)(b−a)=∫f(c)(b - a) = \int ₐᵇ f(x)dxf(x) dx. Solve f(c)=favgf(c) = f_{\mathrm{avg}}.

Worked example 1

For f(x)=xf(x) = x on [0,2][0, 2], find the cc guaranteed by the Mean Value Theorem for Integrals.
  1. f(c)=12∫02x dx=22f(c) = \dfrac{1}{2} \int_{0}^{2} x\,dx = \dfrac{2}{2}
  2. c=22c = \dfrac{2}{2}
  3. c=1∈[0,2]c = 1 \in [0, 2]

Answer: 11

Worked example 2

For f(x)=x2f(x) = x^{2} on [0,3][0, 3], find the cc guaranteed by the Mean Value Theorem for Integrals.
  1. f(c)=13∫03x2 dx=323f(c) = \dfrac{1}{3} \int_{0}^{3} x^{2}\,dx = \dfrac{3^{2}}{3}
  2. c2=323c^{2} = \dfrac{3^{2}}{3}
  3. c=33≈1.73∈[0,3]c = \dfrac{3}{\sqrt{3}} \approx 1.73 \in [0, 3]

Answer: 33≈1.73\dfrac{3}{\sqrt{3}} \approx 1.73

Idea 5 of 5

Net change versus total change

Integrating a signed rate gives net change. Total distance requires splitting where velocity changes sign.

Worked example 1

Velocity is v(t)=t−2v(t) = t - 2 on [0,4][0, 4]. Find displacement.
  1. v(t)=0  ⟹  t=2v(t) = 0 \implies t = 2
  2. ∫02t−2 dt=−42\int_{0}^{2} t - 2\,dt = -\dfrac{4}{2}; ∫24t−2 dt=42\int_{2}^{4} t - 2\,dt = \dfrac{4}{2}
  3. Displacement =−42+42=0= -\dfrac{4}{2} + \dfrac{4}{2} = 0

Answer: 00

Worked example 2

Velocity is v(t)=t−3v(t) = t - 3 on [0,6][0, 6]. Find total distance.
  1. v(t)=0  ⟹  t=3v(t) = 0 \implies t = 3
  2. ∫03t−3 dt=−92\int_{0}^{3} t - 3\,dt = -\dfrac{9}{2}; ∫36t−3 dt=92\int_{3}^{6} t - 3\,dt = \dfrac{9}{2}
  3. Distance =92+92=9= \dfrac{9}{2} + \dfrac{9}{2} = 9

Answer: 99

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §4.4,p4.4, p. 277277. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.