1.4 Integration Formulas and the Net Change Theorem

Big idea

Once a definite integral can be evaluated from any antiderivative, the practical problem changes shape. The hard part is no longer the limit of sums; it is producing a function whose derivative is the integrand. So the first job of this section is stocking a shelf. Every differentiation formula you know, read from right to left, becomes an integration formula, and the short list that results covers a surprising fraction of the integrals you will meet.

Alongside the definite integral there is a notation for the antiderivative itself. The indefinite integral $\int f(x)\,dx$ denotes the whole family of antiderivatives of $f$, which is why it carries a constant. The two objects look alike and are not alike: a definite integral is a number, an indefinite integral is a set of functions. Keeping the constant on one and dropping it on the other is bookkeeping, not decoration.

The second job is interpretation. Reading the evaluation rule with the antiderivative in the starring role gives the net change theorem: integrate a rate of change over an interval and you get the net change in the quantity. Water flowing into a tank at a known rate, a population growing at a known rate, a car moving at a known velocity - in each case the integral of the rate is the total change, and the units multiply out to confirm it.

Net change is where a real distinction appears. Velocity can be negative, and an object that goes out and comes back has a large total distance but a small net displacement. The integral of velocity gives displacement; the integral of speed gives distance. Telling these apart is the single most common source of lost points in this material, and the cure is to find where the velocity changes sign before integrating anything.

Decoder

Integrating a derivative across an interval returns the net change of the original function: $\int_a^b F'(x)\,dx = F(b) - F(a)$.

This is the evaluation rule with the emphasis moved. Read it left to right and it says that if you know how fast something changes at every instant, you can recover how much it changed overall. Read it right to left and it says that a total change can always be recovered by adding up infinitesimal changes. The integrand is a rate and $dx$ supplies the units of the input, so the product carries the units of the output quantity.

The word doing the work is “net”. The two endpoint values are subtracted, so any movement away and back cancels. If a quantity rises by $7$ and then falls by $7$, the integral of its rate is zero over that span even though plenty happened in between.

The fix for the distance question follows from that observation, and it is mechanical. Find the values of $t$ where the velocity is zero, split the interval there, integrate on each piece, and add the absolute values. Integrating $|v(t)|$ is the same instruction stated in symbols, and there is no way to carry it out without first locating the sign changes.

Definitions and results

Indefinite integral. $\int f(x)\,dx = F(x) + C$ means $F'(x) = f(x)$. The constant is part of the answer, because every function differing from $F$ by a constant has the same derivative.

Power rule for antiderivatives. For any exponent $n \ne -1$,

$$ \int x^{n}\,dx = \frac{x^{n+1}}{n+1} + C $$

This covers roots and reciprocal powers once you write them as exponents.

The excluded case. The exponent $-1$ needs its own formula, and the absolute value matters:

$$ \int \frac{1}{x}\,dx = \ln|x| + C $$

Exponential and trigonometric antiderivatives. Read off the derivative rules in reverse:

$$ \int e^{x}\,dx = e^{x} + C, \qquad \int \sin x\,dx = -\cos x + C, \qquad \int \cos x\,dx = \sin x + C $$

$$ \int \sec^2 x\,dx = \tan x + C, \qquad \int \sec x \tan x\,dx = \sec x + C $$

The minus sign on the first trigonometric formula is the one people drop.

Linearity. Constants factor out and sums split, for indefinite and definite integrals alike. There is no product rule and no quotient rule for integrals; a product must be rewritten before it can be integrated term by term.

The net change theorem. If $F'$ is continuous on $[a,b]$ then $\int_a^b F'(x)\,dx = F(b) - F(a)$. Applied to a flow rate in gallons per minute over an interval of minutes, the integral is a number of gallons.

Displacement and distance. For a particle with velocity $v(t)$ on $[a,b]$,

$$ \text{displacement} = \int_a^b v(t)\,dt, \qquad \text{total distance} = \int_a^b |v(t)|\,dt $$

The two agree only when $v$ keeps one sign throughout.

Symmetric intervals. If $f$ is even, meaning $f(-x) = f(x)$, then $\int_{-a}^{a} f = 2\int_0^{a} f$. If $f$ is odd, meaning $f(-x) = -f(x)$, then $\int_{-a}^{a} f = 0$. Spotting symmetry can cut the work in half or remove it entirely.

Worked examples

A sum of powers

Find $\displaystyle\int\left(6x^2 - \frac{4}{x^3} + \frac{2}{\sqrt{x}}\right) dx$.

Rewrite each term as a power first: $6x^2 - 4x^{-3} + 2x^{-1/2}$. Now the power rule applies term by term. The exponents become $3$, $-2$ and $1/2$, so

$$ \int\left(6x^2 - 4x^{-3} + 2x^{-1/2}\right) dx = 2x^3 + 2x^{-2} + 4x^{1/2} + C $$

Check by differentiating: $6x^2$, then $2(-2)x^{-3} = -4x^{-3}$, then $4\cdot\tfrac{1}{2}x^{-1/2} = 2x^{-1/2}$. All three match.

Rewriting before integrating

Find $\displaystyle\int \frac{x^2+3x}{\sqrt{x}}\,dx$.

There is no quotient rule to reverse here, so split the fraction and simplify the exponents:

$$ \frac{x^2+3x}{x^{1/2}} = x^{3/2} + 3x^{1/2} $$

Now integrate term by term:

$$ \int\left(x^{3/2} + 3x^{1/2}\right)dx = \frac{2}{5}x^{5/2} + 2x^{3/2} + C $$

Differentiating gives $\tfrac{2}{5}\cdot\tfrac{5}{2}x^{3/2} + 2\cdot\tfrac{3}{2}x^{1/2} = x^{3/2} + 3x^{1/2}$, the integrand we started from.

Displacement against distance

A particle moves along a line with velocity $v(t) = t^2 - 4t + 3$ metres per second for $0 \le t \le 4$. Find its displacement and the total distance it travels.

Displacement first, since it needs no case analysis. An antiderivative of $v$ is $s(t) = \tfrac{1}{3}t^3 - 2t^2 + 3t$, so

$$ \int_0^4 v(t)\,dt = \left(\frac{64}{3} - 32 + 12\right) - 0 = \frac{64}{3} - 20 = \frac{4}{3} $$

The particle ends up $4/3$ metres to the right of where it started.

For distance, factor the velocity: $v(t) = (t-1)(t-3)$, which is positive on $[0,1)$, negative on $(1,3)$, and positive on $(3,4]$. Integrating on each piece,

$$ \int_0^1 v = \frac{4}{3}, \qquad \int_1^3 v = 0 - \frac{4}{3} = -\frac{4}{3}, \qquad \int_3^4 v = \frac{4}{3} - 0 = \frac{4}{3} $$

Adding absolute values gives a total distance of $4$ metres. Adding the signed values gives $4/3$, the displacement again, which is the check: the signed pieces must reassemble the number computed in one shot.

Using symmetry

Evaluate $\displaystyle\int_{-2}^{2}\left(x^3 + 3x^2\right)dx$.

The first term is odd and the interval is symmetric, so it contributes nothing. The second is even, so it contributes twice its integral over $[0,2]$:

$$ \int_{-2}^{2}\left(x^3+3x^2\right)dx = 0 + 2\int_0^2 3x^2\,dx = 2\big[x^3\big]_0^2 = 16 $$

Direct evaluation confirms it: $\big[\tfrac{1}{4}x^4 + x^3\big]_{-2}^{2} = (4+8) - (4-8) = 12 + 4 = 16$.

Practice

Start with the formulas themselves. These ask for indefinite integrals of powers, exponentials and the standard trigonometric functions, usually after a rewriting step.

Then apply them. These give a rate or a velocity and ask for a net change, a displacement, or a total distance over an interval.

Quiz

Five items on basic antiderivatives, the net change theorem, displacement versus distance, and symmetry.