Integration formulas are the building blocks of calculus, helping us solve complex problems with ease. These formulas, like the Power Rule and Sum Rule, allow us to find antiderivatives and calculate areas under curves.

The Net Change Theorem connects rates of change to total change over time. It's super useful for real-world applications, like finding distance traveled from velocity or population growth from birth rates.

Integration Formulas

Basic integration formulas

• Power Rule applies $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ where $n \neq -1$ to solve problems like $\int x^3 dx = \frac{x^4}{4} + C$
• Constant Multiple Rule multiplies a constant $c$ by the integral of $f(x)$ as $\int cf(x) dx = c \int f(x) dx$ to solve problems like $\int 3x^2 dx = 3 \int x^2 dx = 3 \cdot \frac{x^3}{3} + C = x^3 + C$
• Sum Rule adds the integrals of $f(x)$ and $g(x)$ as $\int [f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx$ to solve problems like $\int (x^2 + 3x) dx = \int x^2 dx + \int 3x dx = \frac{x^3}{3} + \frac{3x^2}{2} + C$
• Difference Rule subtracts the integral of $g(x)$ from the integral of $f(x)$ as $\int [f(x) - g(x)] dx = \int f(x) dx - \int g(x) dx$ to solve problems like $\int (x^2 - 3x) dx = \int x^2 dx - \int 3x dx = \frac{x^3}{3} - \frac{3x^2}{2} + C$

Odd vs even function integrals

• Odd functions satisfy $f(-x) = -f(x)$ and have the property $\int_{-a}^a f(x) dx = 0$ where $a > 0$, so $\int_{-2}^2 x^3 dx = 0$ since $x^3$ is an odd function
• Even functions satisfy $f(-x) = f(x)$ and have the property $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$ where $a > 0$, so $\int_{-1}^1 x^2 dx = 2 \int_0^1 x^2 dx = 2 \cdot \frac{1}{3} = \frac{2}{3}$ since $x^2$ is an even function

Definite integrals and area under a curve

• A definite integral represents the signed area between a function and the x-axis over a specific interval
• For continuous functions, the definite integral can be interpreted as the area under the curve
• The Mean Value Theorem for Integrals states that for a continuous function f(x) on [a,b], there exists a c in [a,b] such that $f(c) = \frac{1}{b-a}\int_a^b f(x)dx$

Net Change Theorem

Net change theorem applications

• States the integral of a rate of change function over $[a, b]$ equals the net change of the quantity over that interval, expressed as $\int_a^b \frac{dQ}{dt} dt = Q(b) - Q(a)$ where $Q(t)$ is the quantity as a function of time $t$
• Applies to scenarios like:
  1. Calculating distance traveled by an object given its velocity function (cars, projectiles)
  2. Determining total change in population given a population growth rate function (bacteria, cities)
  3. Computing accumulated interest earned given an interest rate function (savings accounts, loans)
• Can be used to solve certain differential equations by finding the antiderivative of the rate of change function

Integrals for net change calculations

• Distance traveled formula $\int_a^b v(t) dt = s(b) - s(a)$ uses velocity function $v(t)$ and position function $s(t)$, so if $v(t) = 3t^2$, the distance traveled from $t = 1$ to $t = 3$ is $\int_1^3 3t^2 dt = t^3 \bigg|_1^3 = 27 - 1 = 26$
• Population change formula $\int_a^b r(t) dt = P(b) - P(a)$ uses population growth rate function $r(t)$ and population function $P(t)$, so if $r(t) = 100e^{0.02t}$, the population change from $t = 0$ to $t = 10$ is $\int_0^{10} 100e^{0.02t} dt = 5000e^{0.02t} \bigg|_0^{10} \approx 1103$
• Accumulated interest formula $\int_a^b i(t) dt = A(b) - A(a)$ uses interest rate function $i(t)$ and account balance function $A(t)$, so if $i(t) = 0.05$, the accumulated interest from $t = 0$ to $t = 5$ on an initial balance of $1000 is $\int_0^5 0.05 \cdot 1000 dt = 50t \bigg|_0^5 = 250$