Unit 5, lesson 53

The Fundamental Theorem, Part 2, and Net Change

How to evaluate a definite integral exactly with any antiderivative, which hypotheses make that legal, and how integrating a rate gives net change, displacement and total distance.

  • Theory part 105
  • Practice part 106
  • The Fundamental Theorem of Calculus

Theory, part 105

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Practice, part 106

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The limit of Riemann sums for x2x^2 took real effort to compute, and most functions would be far worse. Part 2 of the fundamental theorem replaces all of that with a subtraction. If you can find an antiderivative, the integral is its change from one end of the interval to the other.

Key terms

Evaluation using an antiderivative

Definition (Fundamental theorem of calculus, part 2.) Let ff be continuous on [a,b][a, b] and let FF be any antiderivative of ff on [a,b][a, b], meaning FF is continuous on [a,b][a, b] and F′(x)=f(x)F'(x) = f(x) for all xx in (a,b)(a, b). Then

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

In plain words To find the integral, find a function whose derivative is the integrand, then subtract its value at the lower bound from its value at the upper bound.

Example The function F(x)=x33F(x) = \frac{x^3}{3} is an antiderivative of x2x^2, so

∫12x2 dx=83−13=73.\int_1^2 x^2 \, dx = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}.

Setting the bounds to 00 and 11 gives 13\frac{1}{3}, the area we found with a page of Riemann sums.

Why the choice of antiderivative does not matter

Definition If FF and GG are both antiderivatives of ff on an interval, then G=F+CG = F + C there for some constant CC, a consequence of the mean value theorem. Hence

G(b)−G(a)=(F(b)+C)−(F(a)+C)=F(b)−F(a).G(b) - G(a) = \big(F(b) + C\big) - \big(F(a) + C\big) = F(b) - F(a).

The difference does not depend on which antiderivative you use.

In plain words The constant of integration cancels in the subtraction, so you can always take C=0C = 0.

Example For ∫12x2 dx\int_1^2 x^2 \, dx try G(x)=x33+10G(x) = \frac{x^3}{3} + 10. Then

G(2)−G(1)=(83+10)−(13+10)=73,G(2) - G(1) = \left( \frac{8}{3} + 10 \right) - \left( \frac{1}{3} + 10 \right) = \frac{7}{3},

the same as before.

Evaluation notation

Definition For a function FF defined at aa and bb we write

[F(x)]ab=F(x)∣ab=F(b)−F(a).\Big[ F(x) \Big]_a^b = F(x) \Big\vert_a^b = F(b) - F(a).

In plain words The bracket is a placeholder that says plug in the top number, plug in the bottom number, subtract.

Example Since −cos⁡x-\cos x is an antiderivative of sin⁡x\sin x,

∫0πsin⁡x dx=[−cos⁡x]0π=(−cos⁡π)−(−cos⁡0)=1+1=2.\int_0^{\pi} \sin x \, dx = \Big[ -\cos x \Big]_0^{\pi} = (-\cos \pi) - (-\cos 0) = 1 + 1 = 2.

So one hump of the sine curve has area exactly 22.

Continuity and antiderivative hypotheses

Definition The theorem needs two things on the whole closed interval [a,b][a, b]. The integrand must be integrable there, which continuity guarantees, and FF must be an antiderivative on the entire interval, continuous on [a,b][a, b] with F′=fF' = f on (a,b)(a, b). More generally the formula holds whenever ff is integrable on [a,b][a, b] and FF is such an antiderivative. If either condition fails at even one point, the formula can give a wrong answer.

In plain words Before subtracting, check that the integrand has no break or blow up inside the interval and that your antiderivative works all the way across.

Example A careless computation gives

∫−111x2 dx=?[−1x]−11=−1−1=−2,\int_{-1}^{1} \frac{1}{x^2} \, dx \overset{?}{=} \Big[ -\frac{1}{x} \Big]_{-1}^{1} = -1 - 1 = -2,

which is absurd, since the integrand is positive. The function 1x2\frac{1}{x^2} is unbounded near 00, so it is not integrable on [−1,1][-1, 1], and −1x-\frac{1}{x} is not even defined at 00. The theorem does not apply.

Net change theorem

Definition If FF is differentiable on [a,b][a, b], with one-sided derivatives at the endpoints, and F′F' is continuous on [a,b][a, b], then

∫abF′(x) dx=F(b)−F(a).\int_a^b F'(x) \, dx = F(b) - F(a).

In plain words Integrating a rate of change gives the total change in the quantity. This is part 2 read the other way around, with FF as the quantity and F′F' as its rate.

Example Water flows into a tank at r(t)=20−tr(t) = 20 - t liters per minute for 0≤t≤100 \le t \le 10. The amount added over those ten minutes is

∫010(20−t) dt=[20t−t22]010=200−50=150 liters.\int_0^{10} (20 - t) \, dt = \Big[ 20t - \frac{t^2}{2} \Big]_0^{10} = 200 - 50 = 150 \text{ liters}.

Total distance versus displacement

Definition Let a particle move along a line with continuous velocity v(t)v(t) on [a,b][a, b] and position s(t)s(t), so s′=vs' = v. Then

displacement=s(b)−s(a)=∫abv(t) dt,total distance=∫ab∣v(t)∣ dt.\text{displacement} = s(b) - s(a) = \int_a^b v(t) \, dt, \qquad \text{total distance} = \int_a^b \lvert v(t) \rvert \, dt.

To compute total distance, split [a,b][a, b] at the times where vv changes sign and add the absolute values of the integrals over the pieces.

In plain words Displacement is where you ended up compared with where you started, and backward motion cancels forward motion. Distance is what the odometer reads, and nothing cancels.

Example Let v(t)=t2−4t+3=(t−1)(t−3)v(t) = t^2 - 4t + 3 = (t - 1)(t - 3) meters per second on [0,4][0, 4], with antiderivative F(t)=t33−2t2+3tF(t) = \frac{t^3}{3} - 2t^2 + 3t.

tt00113344
F(t)F(t)0043\frac{4}{3}0043\frac{4}{3}

The displacement is F(4)−F(0)=43F(4) - F(0) = \frac{4}{3} meters. The velocity is positive on (0,1)(0, 1), negative on (1,3)(1, 3) and positive on (3,4)(3, 4), so the total distance is

43+∣0−43∣+43=4 meters.\frac{4}{3} + \left\lvert 0 - \frac{4}{3} \right\rvert + \frac{4}{3} = 4 \text{ meters}.

Check yourself

  1. Evaluate ∫14(2x+1x)dx\int_1^4 \left( 2x + \frac{1}{\sqrt{x}} \right) dx.
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An antiderivative is x2+2xx^2 + 2\sqrt{x}, so

∫14(2x+1x)dx=(16+4)−(1+2)=17.\int_1^4 \left( 2x + \frac{1}{\sqrt{x}} \right) dx = (16 + 4) - (1 + 2) = 17.
  1. Evaluate ∫0π/2cos⁡x dx\int_0^{\pi/2} \cos x \, dx.
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Since sin⁡x\sin x is an antiderivative of cos⁡x\cos x,

∫0π/2cos⁡x dx=[sin⁡x]0π/2=1−0=1.\int_0^{\pi/2} \cos x \, dx = \Big[ \sin x \Big]_0^{\pi/2} = 1 - 0 = 1.
  1. A cart moves with velocity v(t)=3−tv(t) = 3 - t meters per second for 0≤t≤60 \le t \le 6. Find its displacement and the total distance it travels.
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The displacement is

∫06(3−t) dt=[3t−t22]06=18−18=0.\int_0^6 (3 - t) \, dt = \Big[ 3t - \frac{t^2}{2} \Big]_0^6 = 18 - 18 = 0.

The velocity changes sign at t=3t = 3, and each half contributes 4.54.5 in size, so the distance is 99 meters.

How this lesson connects