1.4 Integration Formulas and the Net Change Theorem

Integration formulas give you a toolkit for finding antiderivatives, which is the core skill you need for everything else in this course. The Net Change Theorem then connects those antiderivatives back to real-world quantities, letting you go from a rate of change to the total accumulated change over an interval.

Integration Formulas

Basic integration formulas

These rules let you break complicated integrals into simpler pieces. You'll use them constantly, so they're worth memorizing cold.

• Power Rule: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ where $n \neq -1$. For example, $\int x^3 \, dx = \frac{x^4}{4} + C$. The restriction $n \neq -1$ matters because plugging in $n = -1$ gives division by zero. (That case gives $\ln|x| + C$ instead.)
• Constant Multiple Rule: You can pull constants out of the integral. $\int c \cdot f(x) \, dx = c \int f(x) \, dx$. For example, $\int 3x^2 \, dx = 3 \int x^2 \, dx = 3 \cdot \frac{x^3}{3} + C = x^3 + C$.
• Sum/Difference Rule: You can split integrals across addition or subtraction. $\int [f(x) \pm g(x)] \, dx = \int f(x) \, dx \pm \int g(x) \, dx$. For example:

$\int (x^2 + 3x) \, dx = \frac{x^3}{3} + \frac{3x^2}{2} + C$

The strategy for most basic integrals is to use the Sum/Difference Rule to separate terms, pull out constants, then apply the Power Rule to each term individually.

Odd vs. even function integrals

When you're integrating over a symmetric interval $[-a, a]$, checking whether your function is odd or even can save you a lot of work.

• Odd functions satisfy $f(-x) = -f(x)$. Their graphs are symmetric about the origin, so the positive and negative areas cancel exactly. This gives you $\int_{-a}^{a} f(x) \, dx = 0$. For example, $\int_{-2}^{2} x^3 \, dx = 0$ because $x^3$ is odd.
• Even functions satisfy $f(-x) = f(x)$. Their graphs are symmetric about the y-axis, so the area from $-a$ to $0$ equals the area from $0$ to $a$. This gives you $\int_{-a}^{a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx$. For example, $\int_{-1}^{1} x^2 \, dx = 2\int_{0}^{1} x^2 \, dx = 2 \cdot \frac{1}{3} = \frac{2}{3}$.

Definite integrals and area under a curve

A definite integral $\int_a^b f(x) \, dx$ represents the signed area between $f(x)$ and the x-axis on $[a, b]$. "Signed" means regions below the x-axis count as negative. If you want total area (all positive), you'd need to integrate $|f(x)|$ instead.

The Mean Value Theorem for Integrals says that for a continuous function on $[a, b]$, there's some point $c$ in $[a, b]$ where:

$f(c) = \frac{1}{b - a}\int_a^b f(x) \, dx$

The right side is the average value of $f$ on $[a, b]$. The theorem guarantees the function actually hits its average value at least once on the interval.

Net Change Theorem

Net change theorem applications

The theorem states that integrating a rate of change over an interval gives you the total (net) change in the original quantity:

$\int_a^b \frac{dQ}{dt} \, dt = Q(b) - Q(a)$

This is really just the Fundamental Theorem of Calculus applied to a specific interpretation. If you know how fast something is changing, integrating that rate tells you how much it changed overall.

Common applications include:

1. Displacement from velocity: Integrating $v(t)$ gives net change in position.
2. Population change from growth rate: Integrating a rate like $r(t)$ (individuals per year) gives total population change.
3. Accumulated cost, revenue, or interest from a rate function over time.

Integrals for net change calculations

Displacement example: If $v(t) = 3t^2$, the displacement from $t = 1$ to $t = 3$ is:

$\int_1^3 3t^2 \, dt = t^3 \Big|_1^3 = 27 - 1 = 26$

One important distinction: $\int_a^b v(t) \, dt$ gives displacement (net change in position), which can be negative if the object reverses direction. To find total distance traveled, you integrate $|v(t)|$ instead, which counts all movement as positive.

Population example: If a population grows at rate $r(t) = 100e^{0.02t}$ individuals per year, the change from $t = 0$ to $t = 10$ is:

$\int_0^{10} 100e^{0.02t} \, dt = \frac{100}{0.02}e^{0.02t}\Big|_0^{10} = 5000(e^{0.2} - 1) \approx 1107$

Interest example: If an account earns at a constant rate of $\text{\textdollar}50$ per year (i.e., $i(t) = 50$), the accumulated interest over 5 years is:

$\int_0^5 50 \, dt = 50t \Big|_0^5 = 250$

In each case, the pattern is the same: identify the rate function, set up the definite integral over your time interval, and evaluate using the Fundamental Theorem.