♾️AP Calculus AB/BC Unit 6 – Integration and Accumulation of Change

Key Concepts and Definitions

• Integration involves finding the area under a curve, which is the opposite of differentiation that finds the slope of a tangent line at a point
• Definite integrals have specific start and end points (bounds) and produce a single value
  • Represented using the notation $\int_a^b f(x) dx$, where $a$ and $b$ are the lower and upper bounds, respectively
• Indefinite integrals lack specific bounds and result in a function plus a constant of integration ($+C$)
  • Represented using the notation $\int f(x) dx$
• Antiderivatives are functions whose derivative is the original function
  • For example, if $F(x)$ is an antiderivative of $f(x)$, then $F'(x) = f(x)$
• Riemann sums approximate the area under a curve by dividing it into rectangles and summing their areas
  • As the number of rectangles approaches infinity, the approximation becomes more accurate
• The average value of a function $f(x)$ over the interval $[a, b]$ is given by $\frac{1}{b-a} \int_a^b f(x) dx$

Fundamental Theorem of Calculus

• The Fundamental Theorem of Calculus (FTC) establishes the relationship between differentiation and integration
• The First Fundamental Theorem of Calculus states that if $F(x)$ is an antiderivative of $f(x)$ on $[a, b]$, then $\int_a^b f(x) dx = F(b) - F(a)$
  • This theorem allows us to evaluate definite integrals using antiderivatives
• The Second Fundamental Theorem of Calculus states that if $f(x)$ is continuous on $[a, b]$, then $\frac{d}{dx} \int_a^x f(t) dt = f(x)$
  • This theorem establishes that differentiation and integration are inverse operations
• The FTC enables us to find the exact area under a curve without using Riemann sums
• The Mean Value Theorem for Integrals is a consequence of the FTC and states that for a continuous function $f(x)$ on $[a, b]$, there exists a point $c \in [a, b]$ such that $\int_a^b f(x) dx = f(c)(b-a)$

Integration Techniques

• Integration by substitution involves changing the variable of integration to simplify the integral
  • Substitute $u = g(x)$, then $du = g'(x) dx$, and rewrite the integral in terms of $u$
• Integration by parts is used when the integrand is a product of two functions, one of which is easier to integrate than the other
  • The formula for integration by parts is $\int u dv = uv - \int v du$
• Trigonometric substitution is used when the integrand contains $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$
  • Substitute using trigonometric functions (sine, cosine, or tangent) to simplify the integral
• Partial fraction decomposition is used when the integrand is a rational function (ratio of polynomials)
  • Decompose the rational function into a sum of simpler fractions, then integrate each term separately
• Improper integrals are integrals with infinite bounds or where the integrand is undefined at one or more points within the bounds
  • Evaluate improper integrals using limits to determine convergence or divergence

Applications of Integration

• Area between curves can be found by integrating the difference between the upper and lower functions
  • $A = \int_a^b [f(x) - g(x)] dx$, where $f(x) \geq g(x)$ on $[a, b]$
• Volume of solids of revolution can be calculated using the disk method or the shell method
  • Disk method: $V = \int_a^b \pi [f(x)]^2 dx$ (revolving around the x-axis)
  • Shell method: $V = \int_a^b 2\pi x f(x) dx$ (revolving around the y-axis)
• Arc length of a curve $y = f(x)$ over the interval $[a, b]$ is given by $L = \int_a^b \sqrt{1 + [f'(x)]^2} dx$
• Work done by a variable force $F(x)$ over a distance $[a, b]$ is calculated using $W = \int_a^b F(x) dx$
• Center of mass of a thin rod with density function $\rho(x)$ over the interval $[a, b]$ is given by $\bar{x} = \frac{\int_a^b x \rho(x) dx}{\int_a^b \rho(x) dx}$

Accumulation Functions

• Accumulation functions represent the accumulated quantity of a rate function over an interval
• If $f(x)$ is a rate function, then the accumulation function $F(x)$ is defined as $F(x) = \int_a^x f(t) dt$
  • $F(x)$ gives the accumulated quantity from $a$ to $x$
• The derivative of an accumulation function is the original rate function: $F'(x) = f(x)$
• The Fundamental Theorem of Calculus connects accumulation functions and definite integrals
  • $F(b) - F(a) = \int_a^b f(x) dx$
• Accumulation functions can be used to solve problems involving total change or accumulated quantities
  • For example, if $f(t)$ represents the rate of change of population over time, then $F(t) = \int_0^t f(x) dx$ gives the total population change from time 0 to time $t$

Solving Differential Equations

• Differential equations involve derivatives of unknown functions and can be solved using integration
• Separable differential equations can be written in the form $\frac{dy}{dx} = f(x)g(y)$
  • Separate variables and integrate both sides: $\int \frac{1}{g(y)} dy = \int f(x) dx$
• Initial value problems (IVPs) are differential equations with a given initial condition
  • Solve the differential equation and use the initial condition to determine the constant of integration
• Exponential growth and decay problems can be modeled using differential equations
  • For exponential growth, $\frac{dy}{dt} = ky$, where $k > 0$
  • For exponential decay, $\frac{dy}{dt} = -ky$, where $k > 0$
• Logistic growth problems involve populations with limited resources and can be modeled using the differential equation $\frac{dy}{dt} = ky(1 - \frac{y}{M})$, where $M$ is the carrying capacity

Common Mistakes and Tips

• Remember to include the constant of integration ($+C$) when finding indefinite integrals
• Be careful with the order of operations when evaluating definite integrals using the Fundamental Theorem of Calculus
• When using integration by parts, choose the term that is easier to integrate as $dv$
• Make sure to use appropriate trigonometric identities when performing trigonometric substitution
• Check for convergence or divergence when evaluating improper integrals
• Pay attention to the units when solving application problems (area, volume, work, etc.)
• Practice various integration techniques to develop fluency and identify the most efficient method for a given problem

Practice Problems and Solutions

1. Evaluate $\int_0^1 (3x^2 + 2x) dx$ Solution: $\int_0^1 (3x^2 + 2x) dx = [x^3 + x^2]_0^1 = (1^3 + 1^2) - (0^3 + 0^2) = 2$

2. Find the area between the curves $y = x^2$ and $y = x + 2$ over the interval $[0, 2]$ Solution: $A = \int_0^2 [(x+2) - x^2] dx = [\frac{x^2}{2} + 2x - \frac{x^3}{3}]_0^2 = (\frac{2^2}{2} + 2 \cdot 2 - \frac{2^3}{3}) - (\frac{0^2}{2} + 2 \cdot 0 - \frac{0^3}{3}) = \frac{10}{3}$

3. Solve the initial value problem $\frac{dy}{dx} = 3x^2$, $y(0) = 1$ Solution: $\int dy = \int 3x^2 dx \rightarrow y = x^3 + C$. Using the initial condition, $1 = 0^3 + C \rightarrow C = 1$. Therefore, the solution is $y = x^3 + 1$

4. Evaluate $\int \frac{x+1}{x^2+2x+1} dx$ using partial fraction decomposition Solution: $\frac{x+1}{x^2+2x+1} = \frac{A}{x+1} + \frac{B}{(x+1)^2}$. Solve for $A$ and $B$ to get $A = 1$ and $B = -1$. Then, $\int \frac{x+1}{x^2+2x+1} dx = \int (\frac{1}{x+1} - \frac{1}{(x+1)^2}) dx = \ln|x+1| + \frac{1}{x+1} + C$

5. Find the volume of the solid generated by revolving the region bounded by $y = \sqrt{x}$, $y = 0$, $x = 1$, and $x = 4$ about the x-axis Solution: Using the disk method, $V = \int_1^4 \pi [\sqrt{x}]^2 dx = \pi \int_1^4 x dx = \pi [\frac{x^2}{2}]_1^4 = \pi (\frac{4^2}{2} - \frac{1^2}{2}) = \frac{15\pi}{2}$