5.4 Integration Formulas and the Net Change Theorem

Integration Formulas

Integration formulas give you a toolkit of known antiderivatives so you don't have to derive everything from scratch. Combined with the Net Change Theorem, they let you turn real-world rate information (speed, flow rate, growth rate) into concrete answers about total change.

Basic Integration Formulas

Power Rule is the one you'll use most often:

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

For example, $\int x^3 \, dx = \frac{x^4}{4} + C$. The restriction $n \neq -1$ matters because plugging in $n = -1$ would put zero in the denominator. That case has its own rule: $\int x^{-1} \, dx = \ln|x| + C$.

Constant Multiple Rule lets you pull constants out front and deal with them separately:

$\int k \cdot f(x) \, dx = k \int f(x) \, dx$

So $\int 3x^2 \, dx = 3 \int x^2 \, dx = 3 \cdot \frac{x^3}{3} + C = x^3 + C$.

Sum and Difference Rules let you split an integral apart term by term:

$\int [f(x) \pm g(x)] \, dx = \int f(x) \, dx \pm \int g(x) \, dx$

For $\int (x^2 + 3x) \, dx$, you integrate each piece: $\frac{x^3}{3} + \frac{3x^2}{2} + C$.

Integrals of Odd vs. Even Functions

These symmetry properties can save you significant work on definite integrals over symmetric intervals $[-a, a]$.

Odd functions satisfy $f(-x) = -f(x)$. Their graphs are symmetric about the origin. Examples include $x^3$, $\sin(x)$, and $x$.

• When you integrate an odd function over $[-a, a]$, the negative side exactly cancels the positive side:

$\int_{-a}^{a} f(x) \, dx = 0$

Even functions satisfy $f(-x) = f(x)$. Their graphs are symmetric about the y-axis. Examples include $x^2$, $\cos(x)$, and $|x|$.

• The left and right halves contribute equally, so you can double the integral over just the right half:

$\int_{-a}^{a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx$

Fundamental Theorem of Calculus and Related Concepts

The Fundamental Theorem of Calculus (FTC) is what makes definite integrals computable without limits of Riemann sums. It has two parts:

• FTC Part 1: If $F(x) = \int_a^x f(t) \, dt$, then $F'(x) = f(x)$. In other words, the derivative of an accumulation function gives back the original function. This confirms that differentiation and integration are inverse operations.
• FTC Part 2: If $F$ is any antiderivative of $f$ on $[a, b]$, then:

$\int_a^b f(x) \, dx = F(b) - F(a)$

This is the evaluation formula you use constantly: find an antiderivative, plug in the bounds, subtract.

A definite integral $\int_a^b f(x)\,dx$ gives a specific number representing the signed area between the curve and the x-axis on $[a, b]$. An indefinite integral $\int f(x)\,dx$ gives a family of antiderivatives (which is why you include $+ C$).

The accumulation function $F(x) = \int_a^x f(t)\,dt$ tracks how much area has built up as you move the upper limit from $a$ to $x$. FTC Part 1 tells you its rate of change at any point equals $f(x)$.

Net Change Theorem

Net Change Theorem Interpretation

The Net Change Theorem is a direct application of FTC Part 2. If $Q(t)$ is a quantity and $Q'(t)$ is its rate of change, then:

$\int_a^b Q'(t) \, dt = Q(b) - Q(a)$

The integral of a rate of change over $[a, b]$ equals the net change in the quantity over that interval.

This applies to any quantity that changes over time:

• If $v(t)$ is velocity, $\int_a^b v(t)\,dt$ gives displacement (net change in position)
• If $P'(t)$ is a population growth rate, $\int_a^b P'(t)\,dt$ gives the net change in population
• If $R(t)$ is a flow rate in liters/min, $\int_a^b R(t)\,dt$ gives the net volume added

Note the word net. If velocity is sometimes negative (moving backward), the integral captures the overall change, not total distance. For total distance, you'd integrate $|v(t)|$ instead.

Applications of the Net Change Theorem

Here's a step-by-step approach for applied problems:

1. Identify the rate function. Determine what quantity is changing and what its rate of change is. For example, a car's velocity $v(t) = 3t^2 + 2t$ m/s.

2. Set up the integral using the theorem. The net change in position from $t = 1$ to $t = 3$ is: $\int_1^3 (3t^2 + 2t) \, dt = s(3) - s(1)$

3. Evaluate the integral. Find the antiderivative and apply the bounds: $\left[ t^3 + t^2 \right]_1^3 = (27 + 9) - (1 + 1) = 36 - 2 = 34$

4. Interpret the result in context. The car's displacement from $t = 1$ to $t = 3$ seconds is 34 meters.

Always check what the problem is asking for. "How far does it travel?" (total distance) and "what is its displacement?" (net change) can give different answers if the object reverses direction.