# Unit 6: Integration and Accumulation of Change

# 1. INTRODUCTION

Integration is the ultimate "big picture" tool. While derivatives focus on the now (instantaneous rate), integration stacks all those tiny moments together to find the total accumulation. It’s the mathematical glow-up that turns a series of snapshots into a full movie. Whether you're tracking a reservoir's water level or calculating the total distance a rocket travels, you're using integration to reconstruct the "whole" from its "parts." Let’s dive into the logic of accumulation and the "crown jewel" of calculus: The Fundamental Theorem.

# 2. ALL KEY CONCEPTS, TERMS, FOUNDATIONAL KNOWLEDGE, and PRINCIPLES

# Key Terms and Definitions

• Antiderivative: A function $F$ is an antiderivative of $f$ on an interval $I$ if $F'(x) = f(x)$ for all $x$ in $I$.
• Indefinite Integral: Represented by $\int f(x) dx$, it denotes the family of all antiderivatives of $f$, typically written as $F(x) + C$.
• Definite Integral: Represented by $\int_a^b f(x) dx$, it represents the signed area between the graph of $f(x)$ and the $x$-axis from $x=a$ to $x=b$.
• Integrand: The function $f(x)$ being integrated.
• Limits of Integration: The values $a$ (lower limit) and $b$ (upper limit) that define the interval of integration.
• Riemann Sum: A method for approximating the total area under a curve by dividing the region into shapes (rectangles or trapezoids).
• Accumulation Function: A function defined by a definite integral with a variable upper limit, such as $F(x) = \int_a^x f(t) dt$.
• The Constant of Integration ($C$): A constant added to an indefinite integral to represent the fact that many functions share the same derivative.
• Integrable: A function is integrable on $[a, b]$ if the limit of its Riemann sum exists as the width of the subintervals approaches zero.

# Foundational Principles

1. The Area Problem: Finding the area of an arbitrary region with curved boundaries.
2. The Distance Problem: Finding total distance when the velocity function is known.
3. Linearity of Integration: The integral of a sum is the sum of the integrals, and constants can be factored out.
4. The Fundamental Theorem of Calculus (FTC): The link between differentiation and integration.
5. Improper Integrals (BC Only): Integrals with infinite limits or integrands that approach infinity within the interval.

# 3. IN-DEPTH EXPLANATION of EVERY CONCEPT and PRINCIPLE

# 3.1 Approximating Area with Riemann Sums (Level 0-20)

Before we can find the "exact" area under a curve $y = f(x)$, we approximate it using $n$ subintervals of width $\Delta x$.

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Given an interval $[a, b]$, the width of each subinterval is:
$$\Delta x = \frac{b - a}{n}$$

We can pick different points within each subinterval to determine the height of our rectangles:

• Left Riemann Sum ($L_n$): Height is evaluated at the left endpoint of each subinterval.
  $$L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x$$
• Right Riemann Sum ($R_n$): Height is evaluated at the right endpoint.
  $$R_n = \sum_{i=1}^{n} f(x_i) \Delta x$$
• Midpoint Riemann Sum ($M_n$): Height is evaluated at the center of each subinterval.
  $$M_n = \sum_{i=1}^{n} f\left(\frac{x_{i-1} + x_i}{2}\right) \Delta x$$

Trapezoidal Rule:
Instead of rectangles, we use trapezoids to better follow the curve. The area of a trapezoid is $\frac{1}{2}(b_1 + b_2)h$. For an integral:
$$T_n = \frac{\Delta x}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + \dots + 2f(x_{n-1}) + f(x_n)]$$

# 3.2 The Formal Definition of the Definite Integral (Level 21-40)

The definite integral is defined as the limit of the Riemann sum as the number of subintervals approaches infinity (and the width approaches zero):
$$\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^) \Delta x$$
where $x_i^$ is any sample point in the $i$-th subinterval.

Properties of Definite Integrals:

1. Order of Integration: $\int_a^b f(x) dx = -\int_b^a f(x) dx$
2. Zero Length Interval: $\int_a^a f(x) dx = 0$
3. Constant Multiple: $\int_a^b k \cdot f(x) dx = k \int_a^b f(x) dx$
4. Sum/Difference: $\int_a^b [f(x) \pm g(x)] dx = \int_a^b f(x) dx \pm \int_a^b g(x) dx$
5. Additivity: $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx$

# 3.3 The Fundamental Theorem of Calculus (FTC) (Level 41-60)

The FTC is divided into two parts that connect the derivative and the integral.

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FTC Part 1 (The Derivative of an Integral):
If $f$ is continuous on $[a, b]$, then the function $g(x) = \int_a^x f(t) dt$ has a derivative given by:
$$\frac{d}{dx} \left[ \int_a^x f(t) dt \right] = f(x)$$
Advanced Chain Rule Version:
$$\frac{d}{dx} \left[ \int_{h(x)}^{g(x)} f(t) dt \right] = f(g(x))g'(x) - f(h(x))h'(x)$$

FTC Part 2 (The Evaluation Theorem):
If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$, then:
$$\int_a^b f(x) dx = F(b) - F(a)$$

# 3.4 Integration by Substitution ($u$-substitution) (Level 61-70)

This is the reverse of the Chain Rule. If we have an integral of the form $\int f(g(x))g'(x) dx$:

1. Let $u = g(x)$.
2. Calculate $du = g'(x) dx$.
3. Substitute $u$ and $du$ into the integral: $\int f(u) du$.
4. Integrate with respect to $u$.
5. For definite integrals, remember to change the limits: $u(a)$ and $u(b)$.

# 3.5 Integration by Parts (BC Only) (Level 71-80)

This is the reverse of the Product Rule.
$$\int u , dv = uv - \int v , du$$
Strategy (LIATE): Choose $u$ based on the priority: Logarithmic, Inverse Trig, Algebraic, Trigonometric, Exponential.

# 3.6 Integration by Linear Partial Fractions (BC Only) (Level 81-90)

When integrating rational functions $\frac{P(x)}{Q(x)}$ where $Q(x)$ is a product of linear factors, we decompose the fraction:
$$\frac{1}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$$
We solve for $A$ and $B$ and integrate the resulting logarithmic terms.

# 3.7 Improper Integrals (BC Only) (Level 91-100)

Improper integrals occur when:

1. The limits are infinite: $\int_a^\infty f(x) dx = \lim_{t \to \infty} \int_a^t f(x) dx$.
2. The integrand has a vertical asymptote (discontinuity) at $x=c \in [a, b]$:
   $$\int_a^b f(x) dx = \lim_{t \to c^-} \int_a^t f(x) dx + \lim_{t \to c^+} \int_t^b f(x) dx$$
   If the limit exists, the integral converges. If not, it diverges.

# 4. EXAMPLES

# Example 1: Left Riemann Sum (Level 5)

Approximate $\int_0^2 x^2 dx$ using a Left Riemann Sum with $n=4$ subintervals.

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Solution:
$\Delta x = \frac{2-0}{4} = 0.5$. Endpoints: $x_0=0, x_1=0.5, x_2=1.0, x_3=1.5, x_4=2.0$.
$L_4 = 0.5 [f(0) + f(0.5) + f(1.0) + f(1.5)]$
$L_4 = 0.5 [0^2 + 0.25 + 1 + 2.25] = 0.5 [3.5] = 1.75$.

# Example 2: Trapezoidal Rule from Table (Level 10)

Given the table of values for velocity $v(t)$:

# Example 3: Basic Indefinite Integral (Level 15)

Evaluate $\int (3x^4 - 5\sqrt{x} + \frac{2}{x}) dx$.
Solution:
$\int 3x^4 dx - \int 5x^{1/2} dx + \int \frac{2}{x} dx$
$= \frac{3x^5}{5} - \frac{5x^{3/2}}{3/2} + 2\ln|x| + C = \frac{3}{5}x^5 - \frac{10}{3}x^{3/2} + 2\ln|x| + C$.

# Example 4: Basic Definite Integral (Level 20)

Evaluate $\int_1^3 (e^x + 1) dx$.
Solution:
$[e^x + x]_1^3 = (e^3 + 3) - (e^1 + 1) = e^3 - e + 2$.

# Example 5: Area under Sine (Level 25)

Find the area under $f(x) = \sin(x)$ from $x=0$ to $x=\pi$.

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

# Example 6: FTC Part 1 - Simple (Level 30)

Find $F'(x)$ if $F(x) = \int_5^x \sqrt{t^3 + 1} dt$.
Solution:
By FTC Part 1, $F'(x) = \sqrt{x^3 + 1}$.

# Example 7: FTC Part 1 - Chain Rule (Level 35)

Find $\frac{d}{dx} \int_0^{x^2} \cos(t) dt$.
Solution:
Let $u = x^2$. Then $\frac{du}{dx} = 2x$.
$\frac{d}{dx} = \cos(x^2) \cdot 2x = 2x\cos(x^2)$.

# Example 8: Average Value of a Function (Level 40)

Find the average value of $f(x) = x^2$ on $[0, 3]$.
Solution:
$f_{avg} = \frac{1}{3-0} \int_0^3 x^2 dx = \frac{1}{3} [\frac{x^3}{3}]_0^3 = \frac{1}{3} [9 - 0] = 3$.

# Example 9: $u$-substitution - Linear (Level 45)

Evaluate $\int (2x+5)^{10} dx$.
Solution:
Let $u = 2x+5$, $du = 2dx \Rightarrow dx = \frac{du}{2}$.
$\int u^{10} \frac{du}{2} = \frac{1}{2} \frac{u^{11}}{11} + C = \frac{(2x+5)^{11}}{22} + C$.

# Example 10: $u$-substitution - Trigonometric (Level 50)

Evaluate $\int \tan(x) dx$.
Solution:
$\int \frac{\sin x}{\cos x} dx$. Let $u = \cos x$, $du = -\sin x dx$.
$\int \frac{-1}{u} du = -\ln|u| + C = -\ln|\cos x| + C = \ln|\sec x| + C$.

# Example 11: Definite Integral with $u$-sub (Level 55)

Evaluate $\int_0^1 x e^{x^2} dx$.
Solution:
Let $u = x^2$, $du = 2x dx$. If $x=0, u=0$. If $x=1, u=1$.
$\frac{1}{2} \int_0^1 e^u du = \frac{1}{2} [e^u]_0^1 = \frac{1}{2}(e - 1)$.

# Example 12: Accumulation Interpretation (Level 60)

A tank contains 50 gallons of oil at $t=0$. Oil is pumped in at $R(t) = 20 + 5\sin(t)$ gal/hr. How much oil is in the tank at $t=2$?
Solution:
$Amount = 50 + \int_0^2 (20 + 5\sin(t)) dt$
$= 50 + [20t - 5\cos(t)]_0^2 = 50 + (40 - 5\cos(2)) - (0 - 5\cos(0))$
$= 50 + 40 - 5\cos(2) + 5 = 95 - 5\cos(2) \approx 97.08$ gallons.

# Example 13: Integration by Parts - Simple (BC Only) (Level 65)

Evaluate $\int x \cos(x) dx$.
Solution:
$u = x, dv = \cos(x) dx \Rightarrow du = dx, v = \sin(x)$.
$\int x \cos(x) dx = x\sin(x) - \int \sin(x) dx = x\sin(x) + \cos(x) + C$.

# Example 14: Integration by Parts - $L$ in LIATE (BC Only) (Level 70)

Evaluate $\int \ln(x) dx$.
Solution:
$u = \ln(x), dv = dx \Rightarrow du = \frac{1}{x} dx, v = x$.
$\int \ln(x) dx = x\ln(x) - \int x \frac{1}{x} dx = x\ln(x) - x + C$.

# Example 15: Tabular Integration by Parts (BC Only) (Level 75)

Evaluate $\int x^2 e^x dx$.
Solution:

# Example 16: Partial Fractions - Distinct Linear (BC Only) (Level 78)

Evaluate $\int \frac{1}{x^2 - 5x + 6} dx$.
Solution:
$\frac{1}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.
$1 = A(x-3) + B(x-2)$.
If $x=3, B=1$. If $x=2, A=-1$.
$\int (\frac{-1}{x-2} + \frac{1}{x-3}) dx = -\ln|x-2| + \ln|x-3| + C = \ln|\frac{x-3}{x-2}| + C$.

# Example 17: Improper Integral - Infinite Limit (BC Only) (Level 80)

Evaluate $\int_1^\infty \frac{1}{x^2} dx$.
Solution:
$\lim_{t \to \infty} \int_1^t x^{-2} dx = \lim_{t \to \infty} [-\frac{1}{x}]1^t = \lim{t \to \infty} (-\frac{1}{t} + 1) = 0 + 1 = 1$. (Converges)

# Example 18: Improper Integral - Discontinuity (BC Only) (Level 82)

Evaluate $\int_0^1 \frac{1}{\sqrt{x}} dx$.
Solution:
Discontinuity at $x=0$.
$\lim_{t \to 0^+} \int_t^1 x^{-1/2} dx = \lim_{t \to 0^+} [2\sqrt{x}]t^1 = \lim{t \to 0^+} (2 - 2\sqrt{t}) = 2$. (Converges)

# Example 19: The $p$-series Integral Test (BC Only) (Level 85)

For what values of $p$ does $\int_1^\infty \frac{1}{x^p} dx$ converge?
Solution:
If $p=1$, $\int_1^\infty \frac{1}{x} dx = [\ln x]_1^\infty = \infty$ (Diverges).
If $p \neq 1$, $\int_1^\infty x^{-p} dx = [\frac{x^{-p+1}}{-p+1}]_1^\infty$.
This limit exists only if $-p+1 < 0 \Rightarrow p > 1$.
Conclusion: Converges for $p > 1$, Diverges for $p \le 1$.

# Example 20: FTC Part 1 with Variable Both Limits (Level 88)

Find $\frac{d}{dx} \int_x^{x^2} \ln(t) dt$.
Solution:
$\frac{d}{dx} [\int_1^{x^2} \ln(t) dt - \int_1^x \ln(t) dt]$
$= \ln(x^2) \cdot (2x) - \ln(x) \cdot (1) = 2x \ln(x^2) - \ln(x) = 4x \ln(x) - \ln(x) = (4x-1)\ln(x)$.

# Example 21: Net Change Theorem (Level 90)

A particle's velocity is $v(t) = t^2 - t - 6$. Find the displacement and total distance traveled on $[1, 4]$.
Solution:
Displacement: $\int_1^4 (t^2 - t - 6) dt = [\frac{t^3}{3} - \frac{t^2}{2} - 6t]_1^4 = (\frac{64}{3} - 8 - 24) - (\frac{1}{3} - \frac{1}{2} - 6) = -4.5$.
Total Distance: Roots of $v(t)$ are $t=3, -2$. Only $t=3$ is in $[1, 4]$.
$\int_1^3 |v(t)| dt + \int_3^4 |v(t)| dt = |\int_1^3 (t^2-t-6) dt| + |\int_3^4 (t^2-t-6) dt$.
Calculation gives $\approx 8.167 + 1.833 = 10$.

# Example 22: Using Symmetry (Level 92)

Evaluate $\int_{-5}^5 (x^3 + \cos x) dx$.
Solution:
$x^3$ is odd, so $\int_{-5}^5 x^3 dx = 0$.
$\cos x$ is even, so $\int_{-5}^5 \cos x dx = 2 \int_0^5 \cos x dx = 2[\sin x]_0^5 = 2\sin(5)$.

# Example 23: Integrating with Absolute Value (Level 93)

Evaluate $\int_0^3 |x-1| dx$.
Solution:
The function splits at $x=1$.
$\int_0^1 -(x-1) dx + \int_1^3 (x-1) dx$
$= [x - \frac{x^2}{2}]_0^1 + [\frac{x^2}{2} - x]_1^3$
$= (1 - 0.5) + (4.5 - 3) - (0.5 - 1) = 0.5 + 1.5 + 0.5 = 2.5$.

# Example 24: Integration by Parts - "The Loop" (BC Only) (Level 95)

Evaluate $\int e^x \sin x dx$.
Solution:
Let $I = \int e^x \sin x dx$.
$u = \sin x, dv = e^x dx \Rightarrow du = \cos x dx, v = e^x$.
$I = e^x \sin x - \int e^x \cos x dx$.
Apply Parts again to $\int e^x \cos x dx$:
$u = \cos x, dv = e^x dx \Rightarrow du = -\sin x dx, v = e^x$.
$I = e^x \sin x - [e^x \cos x - \int e^x (-\sin x) dx]$
$I = e^x \sin x - e^x \cos x - I$.
$2I = e^x(\sin x - \cos x) \Rightarrow I = \frac{e^x(\sin x - \cos x)}{2} + C$.

# Example 25: Improper Integral - Double Infinity (BC Only) (Level 96)

Evaluate $\int_{-\infty}^\infty \frac{1}{1+x^2} dx$.
Solution:
$\int_{-\infty}^0 \dots + \int_0^\infty \dots = \lim_{a \to -\infty} [\arctan x]a^0 + \lim{b \to \infty} [\arctan x]_0^b$
$= (0 - (-\pi/2)) + (\pi/2 - 0) = \pi$.

# Example 26: $u$-substitution with Back-Substitution (Level 97)

Evaluate $\int x \sqrt{x-1} dx$.
Solution:
Let $u = x-1 \Rightarrow x = u+1$ and $dx = du$.
$\int (u+1) \sqrt{u} du = \int (u^{3/2} + u^{1/2}) du$
$= \frac{2}{5}u^{5/2} + \frac{2}{3}u^{3/2} + C = \frac{2}{5}(x-1)^{5/2} + \frac{2}{3}(x-1)^{3/2} + C$.

# Example 27: Finding $C$ with Initial Conditions (Level 98)

If $f'(x) = \frac{1}{x^2+1}$ and $f(0) = 5$, find $f(1)$.
Solution:
$f(x) = \int \frac{1}{x^2+1} dx = \arctan(x) + C$.
$f(0) = \arctan(0) + C = 5 \Rightarrow 0 + C = 5 \Rightarrow C=5$.
$f(1) = \arctan(1) + 5 = \frac{\pi}{4} + 5$.

# Example 28: Complex Partial Fractions (BC Only) (Level 98)

Evaluate $\int \frac{x^2}{(x-1)^3} dx$.
Solution:
Substitute $u = x-1 \Rightarrow x = u+1$.
$\int \frac{(u+1)^2}{u^3} du = \int \frac{u^2+2u+1}{u^3} du = \int (\frac{1}{u} + \frac{2}{u^2} + \frac{1}{u^3}) du$
$= \ln|u| - \frac{2}{u} - \frac{1}{2u^2} + C = \ln|x-1| - \frac{2}{x-1} - \frac{1}{2(x-1)^2} + C$.

# Example 29: Limit of a Riemann Sum to Definite Integral (Level 99)

Express $\lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \sqrt{\frac{i}{n}}$ as a definite integral and evaluate.
Solution:
$\Delta x = 1/n$. Let $x_i = i/n$. The interval is $[0, 1]$.
$\int_0^1 \sqrt{x} dx = [\frac{2}{3}x^{3/2}]_0^1 = 2/3$.

# Example 30: Advanced Accumulation Logic (Level 100)

Let $h(x) = \int_0^{x^2} \sqrt{1+t^3} dt$. Find the linearization of $h(x)$ at $x=1$.
Solution:
$h(1) = \int_0^1 \sqrt{1+t^3} dt$ (Keep as integral or use calculator if allowed).
$h'(x) = \sqrt{1+(x^2)^3} \cdot 2x = 2x\sqrt{1+x^6}$.
$h'(1) = 2(1)\sqrt{1+1} = 2\sqrt{2}$.
Linearization: $L(x) = h(1) + 2\sqrt{2}(x-1)$.

# Example 31: Geometric Integration (Level 10)

Evaluate $\int_{-2}^2 \sqrt{4-x^2} dx$ using geometry.

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Solution:
The graph $y = \sqrt{4-x^2}$ is a semi-circle with radius $r=2$.
Area = $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2^2) = 2\pi$.

# Example 32: Integral of a Piecewise Function (Level 40)

Let $f(x) = 2$ for $x < 1$ and $f(x) = 2x$ for $x \ge 1$. Find $\int_0^2 f(x) dx$.
Solution:
$\int_0^1 2 dx + \int_1^2 2x dx = [2x]_0^1 + [x^2]_1^2 = (2-0) + (4-1) = 5$.

# Example 33: Mean Value Theorem for Integrals (Level 50)

Find $c$ such that $f(c) = f_{avg}$ for $f(x) = 3x^2$ on $[0, 2]$.
Solution:
$f_{avg} = \frac{1}{2} \int_0^2 3x^2 dx = \frac{1}{2} [x^3]_0^2 = 4$.
Set $f(c) = 3c^2 = 4 \Rightarrow c^2 = 4/3 \Rightarrow c = \frac{2}{\sqrt{3}}$ (within $[0, 2]$).

# Example 34: Divergent Improper Integral (BC Only) (Level 70)

Evaluate $\int_1^\infty \frac{1}{x} dx$.
Solution:
$\lim_{t \to \infty} [\ln|x|]1^t = \lim{t \to \infty} (\ln t - 0) = \infty$.
The integral diverges.

# Example 35: Substitution with Trig Identity (Level 85)

Evaluate $\int \cos^3 x dx$.
Solution:
$\int \cos^2 x \cos x dx = \int (1-\sin^2 x) \cos x dx$.
Let $u = \sin x, du = \cos x dx$.
$\int (1-u^2) du = u - \frac{u^3}{3} + C = \sin x - \frac{\sin^3 x}{3} + C$.

This concludes Unit 6. The concepts of accumulation and integration developed here serve as the foundation for calculating volumes, solving differential equations, and analyzing complex series in the remaining units of the AP Calculus curriculum.