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Lesson 83 — Fundamental Theorem of Calculus

FTC Part 1 and Part 2. The bridge between derivative and integral. Leibniz rule for variable limits. Newton and Leibniz, 17th century.

Used in: 3.º ano do EM (17 anos) · Equiv. Math II japonês cap. 6 · Equiv. Klasse 12 alemã

∫abf(x) dx=F(b)−F(a),F′(x)=f(x)\int_a^b f(x)\, dx = F(b) - F(a), \quad F'(x) = f(x)

Fundamental Theorem of Calculus (FTC2): the definite integral of f over [a, b] is simply the difference of the antiderivative F at the endpoints. Unites derivative and integral in a single equation.

Choose your door

Rigorous notation, full derivation, hypotheses

Statement and proofs

FTC — Part 1: derive the integral

"FTC1 states that the derivative of the function defined by an integral with variable upper limit equals the integrand evaluated at the upper limit." — OpenStax Calculus Vol. 1, §5.3

Proof of FTC1. By definition of derivative:

G′(x)=lim⁡h→0G(x+h)−G(x)h=lim⁡h→01h∫xx+hf(t) dt.G'(x) = \lim_{h \to 0} \frac{G(x+h) - G(x)}{h} = \lim_{h \to 0} \frac{1}{h} \int_x^{x+h} f(t)\, dt.

By the Mean Value Theorem for Integrals, there exists chc_h between xx and x+hx+h such that ∫xx+hf(t) dt=f(ch)⋅h\int_x^{x+h} f(t)\, dt = f(c_h) \cdot h. Thus:

G′(x)=lim⁡h→0f(ch).G'(x) = \lim_{h \to 0} f(c_h).

As ch→xc_h \to x when h→0h \to 0 and ff is continuous, f(ch)→f(x)f(c_h) \to f(x). Therefore G′(x)=f(x)G'(x) = f(x). □\square

FTC — Part 2: compute the integral

Proof of FTC2. By FTC1, G(x)=∫axf(t) dtG(x) = \int_a^x f(t)\, dt satisfies G′=fG' = f. Since F′=fF' = f also, F−GF - G has zero derivative on (a,b)(a, b), so F(x)=G(x)+CF(x) = G(x) + C for some constant CC. Then:

F(b)−F(a)=[G(b)+C]−[G(a)+C]=G(b)−G(a)=∫abf−0=∫abf.□F(b) - F(a) = [G(b) + C] - [G(a) + C] = G(b) - G(a) = \int_a^b f - 0 = \int_a^b f. \quad \square

Leibniz Rule (variable limits)

Worked examples

Exercise list

45 exercises · 11 with worked solution (25%)

Application 26Understanding 7Modeling 4Challenge 6Proof 2
  1. Ex. 83.1Application

    Compute ∫032x dx\int_0^3 2x\,dx using FTC2.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.141
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    Antiderivative: F(x)=x2F(x)=x^2. FTC2: F(3)−F(0)=9−0=9F(3)-F(0)=9-0=9. B used F(x)=3xF(x)=3x. C multiplied by 2 unnecessarily. D evaluated only at x=3x=3 without subtracting.
  2. Ex. 83.2Application

    Compute ∫−12x3 dx\int_{-1}^2 x^3\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.142
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    Antiderivative: F(x)=x4/4F(x)=x^4/4. F(2)−F(−1)=16/4−1/4=15/4F(2)-F(-1) = 16/4 - 1/4 = 15/4. B forgot to subtract F(−1)F(-1). C got the sign of (−1)4(-1)^4 wrong. D made an arithmetic error.
  3. Ex. 83.3Application

    Compute ∫0πsin⁡x dx\int_0^\pi \sin x\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.145
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    Antiderivative: F(x)=−cos⁡xF(x)=-\cos x. −cos⁡π−(−cos⁡0)=1+1=2-\cos\pi-(-\cos 0)=1+1=2. B confused with ∫02π\int_0^{2\pi}. C swapped the signs. D confused the integral with the interval.
  4. Ex. 83.4Application

    Compute ∫02ex dx\int_0^2 e^x\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.141-var
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    Antiderivative: F(x)=exF(x)=e^x. e2−e0=e2−1e^2 - e^0 = e^2-1. B forgot to subtract 1. C confused the limit with x=1x=1. D made an error computing e2e^2.
  5. Ex. 83.5ApplicationAnswer key

    Compute ∫02(x2−4x+1) dx\int_0^2 (x^2 - 4x + 1)\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.143
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    Antiderivative: F(x)=x3/3−2x2+xF(x)=x^3/3-2x^2+x. F(2)=8/3−8+2=8/3−6F(2)=8/3-8+2=8/3-6. F(0)=0F(0)=0. Result: 8/3−6=−10/38/3-6=-10/3. B got the sign wrong. C thought the function is even. D made an arithmetic error.
    Show step-by-step (with the why)
    1. Antiderivative: F(x)=x3/3−2x2+xF(x)=x^3/3-2x^2+x.
    2. F(2)=8/3−8+2=8/3−6F(2)=8/3-8+2=8/3-6.
    3. F(0)=0F(0)=0.
    4. Result: 8/3−6=−10/38/3-6=-10/3.
  6. Ex. 83.6ApplicationAnswer key

    Compute ∫1e1x dx\int_1^e \frac{1}{x}\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.146
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    Antiderivative: F(x)=ln⁡xF(x)=\ln x. ln⁡e−ln⁡1=1−0=1\ln e - \ln 1 = 1 - 0 = 1. B confused ∫ex\int e^x with ∫1/x\int 1/x. D thought the integral on [1,e][1,e] is zero.
  7. Ex. 83.7Understanding

    If G(x)=∫0x(t2+1) dtG(x) = \int_0^x (t^2 + 1)\,dt, what is G′(x)G'(x) by FTC1?

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.153
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    FTC1: if G(x)=∫axf(t) dtG(x)=\int_a^x f(t)\,dt, then G′(x)=f(x)G'(x)=f(x). Here f(t)=t2+1f(t)=t^2+1, so G′(x)=x2+1G'(x)=x^2+1. B is the antiderivative (computed the integral, not the derivative). C confuses with FTC2. D ignores the theorem.
  8. Ex. 83.8Application

    Compute ddx∫0x2sin⁡t dt\dfrac{d}{dx}\displaystyle\int_0^{x^2} \sin t\,dt.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.155
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    FTC1 + chain rule: f(t)=sin⁡tf(t)=\sin t, upper limit u=x2u=x^2, u′=2xu'=2x. Result: sin⁡(x2)⋅2x\sin(x^2)\cdot 2x. B forgot the chain rule. C integrated instead of differentiating. D confused sin⁡\sin with cos⁡\cos.
  9. Ex. 83.9Application

    Compute ∫0π/4sec⁡2x dx\int_0^{\pi/4} \sec^2 x\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.147
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    Antiderivative: F(x)=tan⁡xF(x)=\tan x. tan⁡(π/4)−tan⁡0=1−0=1\tan(\pi/4)-\tan 0=1-0=1. B confused the result with the upper limit. C got the value of tan⁡(π/4)\tan(\pi/4) wrong. D thought the integral is zero.
  10. Ex. 83.10Application

    Compute ∫19x dx\int_1^9 \sqrt{x}\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.144
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    Antiderivative: F(x)=(2/3)x3/2F(x)=(2/3)x^{3/2}. F(9)=(2/3)(27)=18F(9)=(2/3)(27)=18, F(1)=(2/3)(1)=2/3F(1)=(2/3)(1)=2/3. Result: 18−2/3=52/318-2/3=52/3. B forgot to subtract F(1)F(1). C computed only F(1)F(1). D got 93/29^{3/2} wrong.
  11. Ex. 83.11ApplicationAnswer key

    Compute ddx∫0x31+t2 dt\dfrac{d}{dx}\displaystyle\int_0^{x^3} \sqrt{1 + t^2}\,dt.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.156
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    FTC1 + chain rule: f(t)=sqrt1+t2f(t)=sqrt{1+t^2}, u=x3u=x^3, u′=3x2u'=3x^2. Result: f(x3)cdot3x2=3x2sqrt1+x6f(x^3)cdot 3x^2 = 3x^2sqrt{1+x^6}. B forgot u′u'. C added a negative sign. D integrated instead of differentiated.
    Show step-by-step (with the why)
    1. Identify f(t)=sqrt1+t2f(t)=sqrt{1+t^2}, u=x3u=x^3, u′=3x2u'=3x^2.
    2. Apply FTC1+chain rule: f(u)cdotu′f(u)cdot u'.
    3. Substitute: sqrt1+(x3)2cdot3x2=3x2sqrt1+x6sqrt{1+(x^3)^2}cdot 3x^2 = 3x^2sqrt{1+x^6}.
  12. Ex. 83.12Application

    Compute ∫0π(cos⁡x+sin⁡x) dx\int_0^\pi (\cos x + \sin x)\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.149
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    Antiderivative: F(x)=sinx−cosxF(x)=sin x - cos x. F(pi)−F(0)=(0−(−1))−(0−1)=1+1=2F(pi)-F(0) = (0-(-1))-(0-1) = 1+1=2. B calculated incorrectly. C swapped the signs. D confused the interval length with the result.
  13. Ex. 83.13UnderstandingAnswer key

    If G(x)=∫axf(t) dtG(x) = \int_a^x f(t)\,dt, what is G′(x)G'(x) by FTC1?

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · conceitual
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    FTC1: if G(x)=intaxf(t),dtG(x)=int_a^x f(t),dt with ff continuous, then G′(x)=f(x)G'(x)=f(x). B confuses with FTC2 formula. C is circular tautology. D only holds if fequiv0fequiv 0.
  14. Ex. 83.14Understanding

    If F′(x)=f(x)F'(x) = f(x), what is the correct expression for ∫abf(x) dx\int_a^b f(x)\,dx by FTC2?

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · conceitual 2
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    FTC2: intabf(x),dx=F(b)−F(a)int_a^b f(x),dx = F(b)-F(a). Order matters. B reversed it. C confuses FF (antiderivative) with ff (integrand). D is the integral of FF.
  15. Ex. 83.15Application

    Compute ddx∫x1t3 dt\dfrac{d}{dx}\displaystyle\int_x^1 t^3\,dt.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.154
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    Lower limit variable: swap the limits and add a negative sign. rac{d}{dx}\int_x^1 t^3\,dt = - rac{d}{dx}\int_1^x t^3\,dt = -x^3. B forgot the sign. C differentiated x3x^3 instead of applying FTC1. D integrated.
  16. Ex. 83.16Application

    Compute \int_0^1 rac{1}{1+x^2}\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.151
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    Antiderivative: F(x)=arctanxF(x)=arctan x. arctan1−arctan0=pi/4−0=pi/4arctan 1 - arctan 0 = pi/4 - 0 = pi/4. B confused with the interval. C confused arctan1arctan 1 with 11. D got the value of arctan1arctan 1 wrong.
  17. Ex. 83.17ApplicationAnswer key

    Compute ∫01(x4−2x2+1) dx\int_0^1 (x^4 - 2x^2 + 1)\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.143-var
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    Antiderivative: F(x)=x5/5−2x3/3+xF(x)=x^5/5-2x^3/3+x. F(1)=1/5−2/3+1=3/15−10/15+15/15=8/15F(1)=1/5-2/3+1=3/15-10/15+15/15=8/15. F(0)=0F(0)=0. B and D arithmetic errors. C ignored negative and fractional terms.
  18. Ex. 83.18Modeling

    With v(t)=t2−4t+3v(t) = t^2 - 4t + 3 m/s, compute the net displacement and total distance from t=0t=0 to t=4t=4.

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    Solve onlineActive Calculus · §4.4 · 4.4.5
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    Displacement: int04v,dt=[t3/3−2t2+3t]04=4/3int_0^4 v,dt=[t^3/3-2t^2+3t]_0^4=4/3. Zeros of v=(t−1)(t−3)v=(t-1)(t-3) at t=1,3t=1,3. Distance: 4/3+4/3+4/3=44/3+4/3+4/3=4 m. B swapped the concepts. C and D computation errors.
    Show step-by-step (with the why)
    1. Compute int04v,dt=[t3/3−2t2+3t]04=4/3int_0^4 v,dt=[t^3/3-2t^2+3t]_0^4=4/3.
    2. Zeros of vv: t=1t=1 and t=3t=3.
    3. Integrals by segments: [0,1]o4/3[0,1] o 4/3, [1,3]o−4/3[1,3] o -4/3, [3,4]o4/3[3,4] o 4/3.
    4. Distance: 4/3+4/3+4/3=44/3+4/3+4/3=4 m.
  19. Ex. 83.19Application

    Compute ddx∫xx2et2 dt\dfrac{d}{dx}\displaystyle\int_x^{x^2} e^{t^2}\,dt.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.157
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    Leibniz rule with both variable limits: f(x2)cdot2x−f(x)cdot1=e(x2)2cdot2x−ex2=ex4cdot2x−ex2f(x^2)cdot 2x - f(x)cdot 1 = e^{(x^2)^2}cdot 2x - e^{x^2} = e^{x^4}cdot 2x - e^{x^2}. B forgot the derivatives of the limits. C and D considered only one limit.
  20. Ex. 83.20Application

    Compute ∫02(2x3−3x2) dx\int_0^2 (2x^3 - 3x^2)\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.143-b
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    Antiderivative: F(x)=x4/2−x3F(x)=x^4/2-x^3. F(2)−F(0)=(8−8)−0=0F(2)-F(0)=(8-8)-0=0. Positive and negative regions cancel. B and C evaluation errors. D got the antiderivative wrong.
  21. Ex. 83.21Modeling

    Marginal cost C′(q)=2q+50C'(q) = 2q + 50 R$/unit. What is the total cost to produce the first 100 units?

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    Solve onlineActive Calculus · §4.4 · 4.4.3
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    int0100(2q+50),dq=[q2+50q]0100=10000+5000=15000int_0^{100}(2q+50),dq=[q^2+50q]_0^{100}=10000+5000=15000. B computed only the fixed cost. C used only the variable cost. D doubled the answer.
  22. Ex. 83.22Application

    Define G(x)=∫1x(2t−1) dtG(x) = \int_1^x (2t - 1)\,dt. What are G(x)G(x) and G(3)G(3)?

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.135
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    G(x)=[t2−t]1x=x2−x−(1−1)=x2−xG(x)=[t^2-t]_1^x=x^2-x-(1-1)=x^2-x. G(3)=9−3=6G(3)=9-3=6. B did not evaluate at t=1t=1. C is just the integrand. D forgot the −x-x.
  23. Ex. 83.23Application

    Knowing that ∫05f dx=12\int_0^5 f\,dx = 12 and ∫02f dx=5\int_0^2 f\,dx = 5, compute ∫25f dx\int_2^5 f\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.137
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    Additivity: int25f=int05f−int02f=12−5=7int_2^5 f = int_0^5 f - int_0^2 f = 12-5=7. B added instead of subtracted. C swapped the order. D multiplied.
  24. Ex. 83.24Challenge

    Compute the area of the region bounded by y=x2−xy = x^2 - x and the xx-axis on [0,1][0, 1].

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    Solve onlineActive Calculus · §4.4 · 4.4.7
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    On [0,1][0,1], x2−xleq0x^2-x leq 0. Area =∣int01(x2−x),dx∣=∣1/3−1/2∣=1/6=|int_0^1(x^2-x),dx|=|1/3-1/2|=1/6. B forgot the absolute value and used only 1/31/3. C is the value without absolute value. D got the denominator wrong.
    Show step-by-step (with the why)
    1. Zeros: x(x−1)=0Rightarrowx=0,1x(x-1)=0Rightarrow x=0,1.
    2. On (0,1)(0,1): f(1/2)=−1/4<0f(1/2)=-1/4<0. Area = absolute value of integral.
    3. int01(x2−x),dx=1/3−1/2=−1/6int_0^1(x^2-x),dx=1/3-1/2=-1/6. Area = 1/61/6.
  25. Ex. 83.25Application

    Compute ∫−22(x2−1) dx\int_{-2}^2 (x^2 - 1)\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.143-c
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    Antiderivative: F(x)=x3/3−xF(x)=x^3/3-x. F(2)−F(−2)=(8/3−2)−(−8/3+2)=8/3−2+8/3−2=16/3−4=4/3F(2)-F(-2)=(8/3-2)-(-8/3+2)=8/3-2+8/3-2=16/3-4=4/3. B thought the function is odd on [−2,2][-2,2]. C made an arithmetic error. D forgot the −x-x.
  26. Ex. 83.26Understanding

    Compute ddx∫0xcos⁡(t2) dt\dfrac{d}{dx}\displaystyle\int_0^x \cos(t^2)\,dt without computing the antiderivative.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.139
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    FTC1 directly: rac{d}{dx}\int_0^x \cos(t^2)\,dt = \cos(x^2). No elementary antiderivative exists for cos(t2)cos(t^2), but FTC1 requires no computation. B swapped coscos for sinsin. C added 2x2x unnecessarily (the upper limit is simply xx). D thought the integral cannot be differentiated.
  27. Ex. 83.27Modeling

    Power P(t)=3+0.5tP(t) = 3 + 0.5t kW. What is the energy in the first 12 hours and the cost at R$ 0.85/kWh?

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    Solve onlineActive Calculus · §4.4 · 4.4.2
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    int012(3+0.5t),dt=[3t+0.25t2]012=36+36=72int_0^{12}(3+0.5t),dt=[3t+0.25t^2]_0^{12}=36+36=72 kWh. Cost: 72imes0.85=61.2072 imes 0.85=61.20. B computed only half. C got the cost wrong. D used t=8t=8.
  28. Ex. 83.28Challenge

    Compute ddx∫sin⁡xcos⁡xt2 dt\dfrac{d}{dx}\displaystyle\int_{\sin x}^{\cos x} t^2\,dt.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.159
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    Leibniz rule: f(t)=t2f(t)=t^2. Upper cosxcos x, lower sinxsin x. Result: cos2xcdot(−sinx)−sin2xcdotcosx=−sinxcosx(cosx+sinx)cos^2 xcdot(-sin x)-sin^2 xcdotcos x = -sin xcos x(cos x+sin x). B and C confuse with identities. D computed only one limit.
    Show step-by-step (with the why)
    1. Upper u=cosxu=cos x, u′=−sinxu'=-sin x. Lower v=sinxv=sin x, v′=cosxv'=cos x.
    2. Leibniz rule: f(u)cdotu′−f(v)cdotv′=cos2x(−sinx)−sin2x(cosx)f(u)cdot u' - f(v)cdot v' = cos^2 x(-sin x)-sin^2 x(cos x).
    3. Factor: −sinxcosx(cosx+sinx)-sin xcos x(cos x+sin x).
  29. Ex. 83.29ChallengeAnswer key

    Compute the average value of f(x)=x2f(x) = x^2 on [0,3][0, 3] and find the cc guaranteed by the Integral Mean Value Theorem.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.161
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    Average value: (1/3)int03x2,dx=(1/3)cdot9=3(1/3)int_0^3 x^2,dx = (1/3)cdot 9 = 3. Mean Value Theorem: c2=3Rightarrowc=sqrt3approx1.73c^2=3Rightarrow c=sqrt{3}approx1.73. B confused cc with the average value. C used the integral without dividing. D got the calculation wrong.
  30. Ex. 83.30ProofAnswer key

    How is FTC2 proven from FTC1 when ff is continuous on [a,b][a,b]?

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    Solve onlineActive Calculus · §4.4 · demonstração
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    By FTC1, G(x)=intaxf,dtG(x)=int_a^x f,dt satisfies G′=f=F′G'=f=F'. Thus (F−G)′=0(F-G)'=0, so F−G=CF-G=C. As G(a)=0G(a)=0, C=F(a)C=F(a). At x=bx=b: intabf=G(b)=F(b)−F(a)int_a^b f=G(b)=F(b)-F(a). A is correct. B incorrect: FTC2 follows from FTC1. C absurd. D incorrect: continuity suffices.
  31. Ex. 83.31Application

    Compute ∫01x2 dx\int_0^1 x^2\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.140
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    Antiderivative: F(x)=x3/3F(x)=x^3/3. F(1)−F(0)=1/3F(1)-F(0)=1/3. B swapped the sign. C used wrong coefficient. D thought the integral is zero.
  32. Ex. 83.32Application

    Compute ∫01ex dx\int_0^1 e^x\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.141-b
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    Antiderivative: F(x)=exF(x)=e^x. F(1)−F(0)=e−1F(1)-F(0)=e-1. B forgot to subtract. C confused the result with the value of e0e^0. D got the sign of subtraction wrong.
  33. Ex. 83.33Understanding

    What is the role of FTC1 compared to FTC2 in computing definite integrals?

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    Solve onlineActive Calculus · §4.4 · conceitual
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    FTC1 states that G(x)=intaxfG(x)=int_a^x f is differentiable, but computing G(b)G(b) directly requires evaluating the integral. FTC2 uses any known antiderivative FF and computes F(b)−F(a)F(b)-F(a) without numerical integration. B and C confuse the roles. D incorrectly restricts scope.
  34. Ex. 83.34Application

    Compute \int_1^2 rac{1}{x}\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.146-b
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    Antiderivative: F(x)=lnxF(x)=ln x. ln2−ln1=ln2ln 2 - ln 1 = ln 2. B confused with the upper limit. C reversed the result. D thought the integral is zero.
  35. Ex. 83.35ChallengeAnswer key

    If h(x)=∫0x2ln⁡(t+1) dth(x) = \int_0^{x^2} \ln(t + 1)\,dt, compute h′(x)h'(x).

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.155-b
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    FTC1 with u=x2u=x^2, u′=2xu'=2x: h′(x)=ln(x2+1)cdot2xh'(x) = ln(x^2+1)cdot 2x. B forgot u′u'. C differentiated ln(x2+1)ln(x^2+1) instead of evaluating at limit. D added +1+1 without justification.
  36. Ex. 83.36Application

    Compute ∫02πcos⁡x dx\int_0^{2\pi} \cos x\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.145-b
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    Antiderivative: F(x)=sinxF(x)=sin x. sin(2pi)−sin(0)=0−0=0sin(2pi)-sin(0)=0-0=0. B got the value of sinsin wrong. C confused with int0piint_0^{pi}. D assigned unwarranted value.
  37. Ex. 83.37Application

    What is the average value of f(x)=x2f(x) = x^2 on [0,3][0, 3]?

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.161-b
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    Average value: rac{1}{3-0}int_0^3 x^2,dx = rac{1}{3}cdotleft[ rac{x^3}{3} ight]_0^3 = rac{1}{3}cdot 9 = 3. B is sqrt3sqrt{3} — the point cc of the MVT, not the average value. C made a calculation error. D used the integral without dividing by interval length.
  38. Ex. 83.38UnderstandingAnswer key

    When computing ∫abf(x) dx\int_a^b f(x)\,dx by FTC2 with two different antiderivatives FF and G=F+CG = F + C, the result:

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · conceitual 3
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    All antiderivatives of ff differ by a constant. If G=F+CG=F+C, then G(b)−G(a)=(F(b)+C)−(F(a)+C)=F(b)−F(a)G(b)-G(a)=(F(b)+C)-(F(a)+C)=F(b)-F(a). The constant cancels: the result is unique regardless of antiderivative chosen. B only true if fequiv0fequiv 0. C confuses the antiderivative with the integrand. D incorrect: result is area with sign.
  39. Ex. 83.39ApplicationAnswer key

    Compute ∫01x dx\int_0^1 x\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.140-b
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    Antiderivative: F(x)=x2/2F(x)=x^2/2. F(1)−F(0)=1/2F(1)-F(0)=1/2. B thought the area is zero. C used the primitive without dividing by 2. D swapped the sign.
  40. Ex. 83.40Challenge

    Compute ddx∫0x2cos⁡(t2) dt\dfrac{d}{dx}\displaystyle\int_0^{x^2} \cos(t^2)\,dt.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.155-c
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    FTC1 with u=x2u=x^2, integrand f(t)=cos(t2)f(t)=cos(t^2): f(u)cdotu′=cos((x2)2)cdot2x=cos(x4)cdot2xf(u)cdot u' = cos((x^2)^2)cdot 2x = cos(x^4)cdot 2x. B forgot u′u'. C evaluated the integrand incorrectly. D got the function wrong.
  41. Ex. 83.41Modeling

    The flow rate of water into a tank is r(t)=100tr(t) = 100t L/h. How many liters enter in the first 6 hours?

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    Solve onlineActive Calculus · §4.4 · 4.4.1
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    int06100t,dt=left[50t2ight]06=50cdot36=1800int_0^6 100t,dt=left[50t^2 ight]_0^6=50cdot 36=1800 L. B computed only half the time. C doubled without justification. D used only t=1t=1 hour.
  42. Ex. 83.42Application

    Compute ∫−112x2 dx\int_{-1}^1 2x^2\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.142-b
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    Antiderivative: F(x)=2x3/3F(x)=2x^3/3. F(1)−F(−1)=2/3−(−2/3)=4/3F(1)-F(-1)=2/3-(-2/3)=4/3. B used the antiderivative of x2x^2 without the factor 2. C ignored correct symmetry. D computed only one side.
  43. Ex. 83.43Understanding

    What is the sign of ∫π2πsin⁡x dx\int_\pi^{2\pi} \sin x\,dx?

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    Solve onlineActive Calculus · §4.4 · conceitual 2
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    On (pi,2pi)(pi, 2pi), sinx<0sin x < 0. Thus intpi2pisinx,dx=[−cosx]pi2pi=−cos2pi+cospi=−1+(−1)=−2<0int_{pi}^{2pi}sin x,dx=[-cos x]_{pi}^{2pi}=-cos 2pi+cospi = -1+(-1)=-2<0. B and C get the sign wrong. D ignores the interval.
  44. Ex. 83.44Challenge

    Compute \int_0^1 rac{1}{\sqrt{1-x^2}}\,dx.

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    Solve onlineOpenStax Calculus Volume 1 · §5.3 · 5.3.152
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    Antiderivative: F(x)=arcsinxF(x)=arcsin x. F(1)−F(0)=pi/2−0=pi/2F(1)-F(0)=pi/2-0=pi/2. B doubled. C used the wrong value for arcsin1arcsin 1. D confused with arctan1arctan 1.
  45. Ex. 83.45ProofAnswer key

    The proof of FTC1 uses which essential argument?

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    Solve onlineActive Calculus · §4.4 · demonstração 2
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    The proof uses the Integral MVT: (G(x+h)−G(x))/h=(1/h)intxx+hf=f(ch)(G(x+h)-G(x))/h = (1/h)int_x^{x+h}f = f(c_h) for some chc_h between xx and x+hx+h. When ho0h o 0, choxc_h o x, and by continuity of ff, f(ch)of(x)f(c_h) o f(x). B incorrect. C only continuity, not differentiability, is needed. D incorrect.

Sources

Updated on 2026-05-06 · Author(s): Clube da Matemática

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