# UNIT 8: APPLICATIONS OF INTEGRATION

# 1. INTRODUCTION

If the derivative is the mathematical tool for understanding "instantaneous change," then the integral is the tool for "total accumulation." In the previous units of AP Calculus, we focused on the mechanics of integration—antiderivatives, the Fundamental Theorem of Calculus, and techniques like $u$-substitution and integration by parts. Now, we transition from the how to the why.

Unit 8, Applications of Integration, represents the culmination of integral calculus. It allows us to step out of the abstract world of $x$ and $y$ and into the three-dimensional physical world. How do we calculate the exact volume of a biological cell with an irregular shape? How can an engineer determine the amount of material needed to create a cooling tower for a power plant? How do we find the exact distance an object travels along a curved path?

This chapter bridges the gap between geometry and calculus. We will move beyond simple formulas for the area of a rectangle or the volume of a sphere to find the area and volume of virtually any shape defined by functions. For BC students, we will extend these concepts to the realm of polar coordinates and parametric equations. By the end of this unit, you will see the world not as a collection of static shapes, but as a series of accumulations that can be precisely measured and analyzed.

# 2. KEY CONCEPTS, TERMS, AND PRINCIPLES

# Foundational Terms

• Accumulation Function: A function defined by a definite integral, representing the net change or total amount of a quantity accumulated over an interval.
• Average Value: The representative height of a function over a closed interval $[a, b]$, calculated as $\frac{1}{b-a} \int_a^b f(x) , dx$.
• Area Under a Curve: The geometric interpretation of the definite integral of a non-negative function.
• Area Between Curves: The region bounded by two or more functions, found by integrating the difference between the "top" and "bottom" functions.
• Cross Section: A two-dimensional "slice" of a three-dimensional solid.
• Solid of Revolution: A 3D shape formed by rotating a 2D region around a fixed line (the axis of revolution).
• Disk Method: A technique for finding the volume of a solid of revolution when the region is flush against the axis of revolution.
• Washer Method: A technique for finding the volume of a solid of revolution when there is a gap between the region and the axis of revolution, resulting in a "hole."
• Arc Length (BC Only): The distance along a curved path between two points.
• Polar Area (BC Only): The area swept out by a radius vector in a polar coordinate system.

# Core Principles

• The Net Change Theorem: The integral of a rate of change is the net change: $\int_a^b F'(x) , dx = F(b) - F(a)$.
• The Mean Value Theorem for Integrals: If $f$ is continuous on $[a, b]$, there exists a number $c$ in $[a, b]$ such that $f(c) = \text{Average Value}$.
• Riemann Sum Limit: All applications in this unit are conceptually derived from the limit of a Riemann sum: $\lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x = \int_a^b f(x) , dx$.

# 3. IN-DEPTH EXPLANATION OF CONCEPTS

# 3.1 The Average Value of a Function

In discrete math, the average of $n$ numbers is $\frac{1}{n} \sum x_i$. In calculus, a function $f(x)$ has infinitely many values on $[a, b]$. To find the average, we divide the total accumulation (the integral) by the length of the interval.

The Formula:
$$f_{avg} = \frac{1}{b-a} \int_a^b f(x) , dx$$

The Mean Value Theorem for Integrals:
This theorem guarantees that for a continuous function, there is at least one point $x = c$ where the function actually equals its average value.
$$f(c)(b-a) = \int_a^b f(x) , dx$$
Geometrically, this means there is a rectangle with height $f(c)$ and width $(b-a)$ that has the exact same area as the region under the curve.

# 3.2 Area Between Curves

To find the area between two curves $f(x)$ and $g(x)$, we sum up an infinite number of infinitesimal rectangles.

Integration with respect to $x$:
If $f(x) \ge g(x)$ for all $x$ in $[a, b]$:
$$\text{Area} = \int_a^b [f(x) - g(x)] , dx$$
Tip: Think "Top function minus Bottom function".

Integration with respect to $y$:
Sometimes curves are defined as $x = f(y)$. If $f(y) \ge g(y)$ (meaning $f(y)$ is further to the right):
$$\text{Area} = \int_c^d [f(y) - g(y)] , dy$$
Tip: Think "Right function minus Left function".

# 3.3 Volume with Known Cross Sections

If we know the area of a cross-section $A(x)$ perpendicular to the $x$-axis, the volume $V$ is the integral of that area.

The Formula:
$$V = \int_a^b A(x) , dx$$

Common Cross-Section Areas $A(x)$ based on a base length $s$ (where $s = f(x) - g(x)$):

1. Squares: $A(x) = s^2$
2. Semicircles: $A(x) = \frac{1}{2} \pi (\frac{s}{2})^2 = \frac{\pi}{8} s^2$
3. Equilateral Triangles: $A(x) = \frac{\sqrt{3}}{4} s^2$
4. Isosceles Right Triangle (Leg on base): $A(x) = \frac{1}{2} s^2$
5. Isosceles Right Triangle (Hypotenuse on base): $A(x) = \frac{1}{4} s^2$

# 3.4 Volume of Revolution: Disk and Washer Methods

When a region is rotated around an axis, it creates a solid.

# The Disk Method

Used when the region is adjacent to the axis of rotation. The cross-section is a circle with area $\pi R^2$.
$$V = \pi \int_a^b [R(x)]^2 , dx$$
Where $R(x)$ is the distance from the axis to the function.

# The Washer Method

Used when there is a "gap" between the region and the axis. The cross-section is a ring (washer) with area $\pi(R_{outer}^2 - R_{inner}^2)$.
$$V = \pi \int_a^b ([R(x)]^2 - [r(x)]^2) , dx$$
Where $R(x)$ is the outer radius and $r(x)$ is the inner radius.

# 3.5 Arc Length (BC Only)

To find the length of a smooth curve $y = f(x)$ from $x=a$ to $x=b$, we use the Pythagorean-based formula:
$$L = \int_a^b \sqrt{1 + [f'(x)]^2} , dx$$

For Parametric Equations $x=x(t), y=y(t)$:
$$L = \int_{t_1}^{t_2} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} , dt$$

# 3.6 Area in Polar Coordinates (BC Only)

In polar coordinates, we don't use rectangles; we use circular sectors. The area of a sector is $\frac{1}{2} r^2 \theta$.
The Formula:
$$\text{Area} = \int_\alpha^\beta \frac{1}{2} [r(\theta)]^2 , d\theta$$

For the area between two polar curves $r_{out}$ and $r_{in}$:
$$\text{Area} = \frac{1}{2} \int_\alpha^\beta ([r_{out}(\theta)]^2 - [r_{in}(\theta)]^2) , d\theta$$

# 4. EXAMPLES

# Example 1: Average Value (Level 10)

Find the average value of $f(x) = x^2$ on the interval $[0, 3]$.
Solution:
$$f_{avg} = \frac{1}{3-0} \int_0^3 x^2 , dx = \frac{1}{3} \left[ \frac{x^3}{3} \right]_0^3 = \frac{1}{3} (\frac{27}{3} - 0) = 3$$

# Example 2: Mean Value Theorem for Integrals (Level 20)

Find the value of $c$ guaranteed by the MVT for Integrals for $f(x) = \sqrt{x}$ on $[0, 4]$.
Solution:
First, find the average value:
$$f_{avg} = \frac{1}{4-0} \int_0^4 x^{1/2} , dx = \frac{1}{4} \left[ \frac{2}{3}x^{3/2} \right]_0^4 = \frac{1}{4} (\frac{2}{3} \cdot 8) = \frac{4}{3}$$
Set $f(c) = \frac{4}{3}$:
$$\sqrt{c} = \frac{4}{3} \implies c = \frac{16}{9}$$
Since $16/9 \approx 1.77$ is in $[0, 4]$, $c = 16/9$.

# Example 3: Area Between Curves (Level 15)

Find the area of the region bounded by $f(x) = x^2$ and $g(x) = x + 2$.
Solution:
Find intersections: $x^2 = x + 2 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1)=0$. Intersects at $x = -1, 2$.
On $[-1, 2]$, $x+2 \ge x^2$.
$$\text{Area} = \int_{-1}^2 (x + 2 - x^2) , dx = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^2$$
$$= (2 + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3}) = \frac{10}{3} - (-\frac{7}{6}) = \frac{20+7}{6} = \frac{27}{6} = 4.5$$

# Example 4: Area Integrating with Respect to $y$ (Level 30)

Find the area bounded by $y^2 = x$ and $y = x - 2$.
Solution:
Rewrite as $x = y^2$ and $x = y + 2$.
Intersections: $y^2 = y + 2 \implies y^2 - y - 2 = 0 \implies y = 2, -1$.
Right function: $x = y + 2$. Left function: $x = y^2$.
$$\text{Area} = \int_{-1}^2 (y + 2 - y^2) , dy = \left[ \frac{y^2}{2} + 2y - \frac{y^3}{3} \right]_{-1}^2 = 4.5$$

# Example 5: Cross Sections - Squares (Level 35)

The base of a solid is the circle $x^2 + y^2 = 1$. Cross sections perpendicular to the $x$-axis are squares. Find the volume.
Solution:
The circle is $y = \pm \sqrt{1-x^2}$. The side length $s$ of the square is the distance between top and bottom:
$s = \sqrt{1-x^2} - (-\sqrt{1-x^2}) = 2\sqrt{1-x^2}$.
$A(x) = s^2 = (2\sqrt{1-x^2})^2 = 4(1-x^2)$.
$$V = \int_{-1}^1 4(1-x^2) , dx = 4 [x - \frac{x^3}{3}]_{-1}^1 = 4[(1 - 1/3) - (-1 + 1/3)] = 4[2/3 + 2/3] = \frac{16}{3}$$

# Example 6: Disk Method - X-Axis (Level 25)

Find the volume of the solid generated by rotating $y = \sqrt{x}$ from $x=0$ to $x=4$ about the $x$-axis.
Solution:
$R(x) = \sqrt{x}$.
$$V = \pi \int_0^4 (\sqrt{x})^2 , dx = \pi \int_0^4 x , dx = \pi [\frac{x^2}{2}]_0^4 = 8\pi$$

# Example 7: Washer Method - X-Axis (Level 40)

Find the volume rotating the region bounded by $y = x^2$ and $y = \sqrt{x}$ about the $x$-axis.
Solution:
Intersections at $x=0, 1$. $R_{out} = \sqrt{x}$, $r_{in} = x^2$.
$$V = \pi \int_0^1 ((\sqrt{x})^2 - (x^2)^2) , dx = \pi \int_0^1 (x - x^4) , dx = \pi [\frac{x^2}{2} - \frac{x^5}{5}]_0^1 = \pi (1/2 - 1/5) = \frac{3\pi}{10}$$

# Example 8: Disk Method - Y-Axis (Level 35)

Rotate $y = x^3$, $y = 8$, and $x = 0$ about the $y$-axis.
Solution:
Rewrite as $x = y^{1/3}$. Bounds for $y$ are $0$ to $8$.
$$V = \pi \int_0^8 (y^{1/3})^2 , dy = \pi \int_0^8 y^{2/3} , dy = \pi [\frac{3}{5}y^{5/3}]_0^8 = \pi \cdot \frac{3}{5} \cdot 32 = \frac{96\pi}{5}$$

# Example 9: Washer Method - Off-Axis (Level 50)

Rotate the region bounded by $y = x^2$ and $y = 4$ about the line $y = 5$.
Solution:
Intersections at $x = -2, 2$.
$R_{out} = \text{distance from } y=5 \text{ to } y=x^2 \implies 5 - x^2$.
$r_{in} = \text{distance from } y=5 \text{ to } y=4 \implies 5 - 4 = 1$.
$$V = \pi \int_{-2}^2 ((5-x^2)^2 - 1^2) , dx = \pi \int_{-2}^2 (25 - 10x^2 + x^4 - 1) , dx$$
$$= \pi \int_{-2}^2 (24 - 10x^2 + x^4) , dx = \pi [24x - \frac{10x^3}{3} + \frac{x^5}{5}]_{-2}^2 = \frac{832\pi}{15}$$

# Example 10: Arc Length of a Function (BC Level 45)

Find the arc length of $f(x) = \frac{2}{3}x^{3/2}$ from $x=0$ to $x=3$.
Solution:
$f'(x) = x^{1/2}$.
$$L = \int_0^3 \sqrt{1 + (x^{1/2})^2} , dx = \int_0^3 \sqrt{1 + x} , dx$$
Let $u = 1+x, du=dx$.
$$L = \int_1^4 u^{1/2} , du = [\frac{2}{3}u^{3/2}]_1^4 = \frac{2}{3}(8 - 1) = \frac{14}{3}$$

# Example 11: Arc Length of Parametric Curve (BC Level 50)

Find the length of the path $x = \cos t, y = \sin t$ for $0 \le t \le \pi$.
Solution:
$\frac{dx}{dt} = -\sin t, \frac{dy}{dt} = \cos t$.
$$L = \int_0^\pi \sqrt{(-\sin t)^2 + (\cos t)^2} , dt = \int_0^\pi \sqrt{1} , dt = [t]_0^\pi = \pi$$
(This makes sense as it is half the circumference of a unit circle).

# Example 12: Polar Area - Simple Circle (BC Level 45)

Find the area of $r = 3 \sin \theta$.
Solution:
This is a circle of diameter 3. It is traced once from $0$ to $\pi$.
$$\text{Area} = \frac{1}{2} \int_0^\pi (3 \sin \theta)^2 , d\theta = \frac{9}{2} \int_0^\pi \sin^2 \theta , d\theta$$
Using $\sin^2 \theta = \frac{1-\cos 2\theta}{2}$:
$$\text{Area} = \frac{9}{4} \int_0^\pi (1 - \cos 2\theta) , d\theta = \frac{9}{4} [\theta - \frac{1}{2}\sin 2\theta]_0^\pi = \frac{9\pi}{4}$$

# Example 13: Area of a Cardioid (BC Level 55)

Find the area inside $r = 1 + \cos \theta$.
Solution:
Integrated from $0$ to $2\pi$.
$$\text{Area} = \frac{1}{2} \int_0^{2\pi} (1 + \cos \theta)^2 , d\theta = \frac{1}{2} \int_0^{2\pi} (1 + 2\cos \theta + \cos^2 \theta) , d\theta$$
$$= \frac{1}{2} \int_0^{2\pi} (1 + 2\cos \theta + \frac{1 + \cos 2\theta}{2}) , d\theta = \frac{1}{2} \int_0^{2\pi} (1.5 + 2\cos \theta + 0.5\cos 2\theta) , d\theta$$
$$= \frac{1}{2} [1.5\theta + 2\sin \theta + 0.25\sin 2\theta]_0^{2\pi} = \frac{1}{2} (3\pi) = \frac{3\pi}{2}$$

# Example 14: Net Change - Water Flow (Level 20)

Water flows into a tank at a rate $R(t) = 20e^{-0.1t}$ liters/min. How much water enters from $t=0$ to $t=10$?
Solution:
$$\text{Net Change} = \int_0^{10} 20e^{-0.1t} , dt = 20 [\frac{e^{-0.1t}}{-0.1}]_0^{10} = -200 [e^{-1} - e^0] = 200(1 - \frac{1}{e}) \approx 126.42 \text{ liters.}$$

# Example 15: Cross Sections - Semicircles (Level 45)

Base is bounded by $y = e^x, y=0, x=0, x=ln(3)$. Cross sections perpendicular to the $x$-axis are semicircles.
Solution:
$s = e^x$. Area $A(x) = \frac{\pi}{8}s^2 = \frac{\pi}{8}e^{2x}$.
$$V = \frac{\pi}{8} \int_0^{\ln 3} e^{2x} , dx = \frac{\pi}{8} [\frac{1}{2}e^{2x}]_0^{\ln 3} = \frac{\pi}{16} (e^{2\ln 3} - e^0) = \frac{\pi}{16} (9 - 1) = \frac{\pi}{2}$$

# Example 16: Volume Rotating about $x = k$ (Level 55)

Rotate the region bounded by $x = y^2$ and $x = 1$ about the line $x = 1$.
Solution:
Since the axis is vertical ($x=1$), we integrate with respect to $y$.
Intersections: $y^2 = 1 \implies y = \pm 1$.
The radius is the distance from $x=1$ to $x=y^2$: $R(y) = 1 - y^2$.
$$V = \pi \int_{-1}^1 (1-y^2)^2 , dy = \pi \int_{-1}^1 (1 - 2y^2 + y^4) , dy = \pi [y - \frac{2y^3}{3} + \frac{y^5}{5}]_{-1}^1$$
$$= \pi [(1 - 2/3 + 1/5) - (-1 + 2/3 - 1/5)] = \pi [8/15 + 8/15] = \frac{16\pi}{15}$$

# Example 17: Accumulation Function - Displacement (Level 25)

A particle's velocity is $v(t) = t^2 - 4$. Find the displacement from $t=0$ to $t=3$.
Solution:
$$\text{Displacement} = \int_0^3 (t^2 - 4) , dt = [\frac{t^3}{3} - 4t]_0^3 = (9 - 12) - 0 = -3$$

# Example 18: Total Distance Traveled (Level 40)

Using $v(t) = t^2 - 4$ from Example 17, find total distance on $[0, 3]$.
Solution:
$\text{Distance} = \int_0^3 |t^2 - 4| , dt$.
$t^2 - 4 = 0$ at $t=2$. On $[0, 2]$, $v(t) \le 0$. On $[2, 3]$, $v(t) \ge 0$.
$$\text{Dist} = \int_0^2 -(t^2 - 4) , dt + \int_2^3 (t^2 - 4) , dt$$
$$= [-t^3/3 + 4t]_0^2 + [t^3/3 - 4t]_2^3 = (-8/3 + 8) + [(9-12) - (8/3 - 8)] = \frac{16}{3} + [-3 + 16/3] = \frac{16+7}{3} = \frac{23}{3}$$

# Example 19: Polar Intersection (BC Level 75)

Find the area inside $r = 3 \sin \theta$ and outside $r = 2 - \sin \theta$.
Solution:
Find intersections: $3 \sin \theta = 2 - \sin \theta \implies 4 \sin \theta = 2 \implies \sin \theta = 1/2$.
$\theta = \pi/6, 5\pi/6$.
$$\text{Area} = \frac{1}{2} \int_{\pi/6}^{5\pi/6} [(3 \sin \theta)^2 - (2 - \sin \theta)^2] , d\theta$$
$$= \frac{1}{2} \int_{\pi/6}^{5\pi/6} [9 \sin^2 \theta - (4 - 4 \sin \theta + \sin^2 \theta)] , d\theta$$
$$= \frac{1}{2} \int_{\pi/6}^{5\pi/6} [8 \sin^2 \theta + 4 \sin \theta - 4] , d\theta$$
Using $8 \sin^2 \theta = 4(1 - \cos 2\theta)$:
$$= \frac{1}{2} \int_{\pi/6}^{5\pi/6} [4 - 4\cos 2\theta + 4\sin \theta - 4] , d\theta = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (4\sin \theta - 4\cos 2\theta) , d\theta$$
$$= \frac{1}{2} [-4\cos \theta - 2\sin 2\theta]{\pi/6}^{5\pi/6} = 2 [-\cos \theta - \frac{1}{2}\sin 2\theta]{\pi/6}^{5\pi/6}$$
$$= 2 [(\frac{\sqrt{3}}{2} - \frac{1}{2}(-\frac{\sqrt{3}}{2})) - (-\frac{\sqrt{3}}{2} - \frac{1}{2}(\frac{\sqrt{3}}{2}))] = 3\sqrt{3}$$

# Example 20: Work as an Integral (Level 60 - Contextual)

A cable weighing 2 lbs/ft is used to lift a 100 lb bucket from a 50 ft well. Find the work done.
Solution:
Work = $\int F(x) , dx$. Let $x$ be the length of cable already pulled up.
Weight at height $x$ is: (Bucket) + (Remaining Cable) $= 100 + 2(50 - x) = 200 - 2x$.
$$W = \int_0^{50} (200 - 2x) , dx = [200x - x^2]_0^{50} = 10000 - 2500 = 7500 \text{ ft-lb.}$$

# Example 21: Area between curves (Integrating x vs y) (Level 50)

Find the area of the region in the first quadrant bounded by $y = \sqrt{x}$, $y = x-2$, and the $y$-axis.
Solution:
This region has two "bottom" functions if integrating $dx$ (from $0$ to $2$, bottom is $y=0$; from $2$ to $4$, bottom is $x-2$). It is easier to use $dy$.
Left curve: $x = 0$. Right curve: $y = \sqrt{x} \implies x = y^2$ (initially) then $y = x-2 \implies x = y+2$.
Actually, the right boundaries are $x = y^2$ and the $y$-axis is the left.
Wait, let's re-sketch. The curves are $y = \sqrt{x}$ (top), $x = 0$ (left), $y=x-2$ (right/bottom).
Intersection: $\sqrt{x} = x-2 \implies x = x^2 - 4x + 4 \implies x^2 - 5x + 4 = 0 \implies x=4, y=2$.
Integrating $dy$ from $y=0$ to $y=2$:
Right curve is $x = y+2$. Left curve is $x = y^2$.
$$\text{Area} = \int_0^2 (y+2 - y^2) , dy = [\frac{y^2}{2} + 2y - \frac{y^3}{3}]_0^2 = 2 + 4 - 8/3 = 10/3$$

# Example 22: Volume by Rotation - Shell-like setup with Washer (Level 70)

Region bounded by $y = e^{-x^2}, y=0, x=0, x=1$. Rotate about $x = -1$.
Solution:
(Note: AP Calculus usually uses Washer/Disk. While Shell is a BC topic in some schools, for AP we stick to Washer by integrating $dy$ or we set up the integral based on the geometry).
Since rotating about a vertical axis $x=-1$, and the function is $y=f(x)$, Shell Method is most efficient:
$V = 2\pi \int_a^b (\text{radius})(\text{height}) , dx$
Radius $= x - (-1) = x+1$. Height $= e^{-x^2}$.
$$V = 2\pi \int_0^1 (x+1)e^{-x^2} , dx = 2\pi \left[ \int_0^1 xe^{-x^2} dx + \int_0^1 e^{-x^2} dx \right]$$
The second part is non-elementary, so this would be a calculator-active question on the AP exam.

# Example 23: Arc Length (Function of $y$) (BC Level 60)

Find the length of $x = \frac{1}{3}(y^2 + 2)^{3/2}$ from $y=0$ to $y=1$.
Solution:
$\frac{dx}{dy} = \frac{1}{3} \cdot \frac{3}{2}(y^2 + 2)^{1/2} \cdot 2y = y\sqrt{y^2 + 2}$.
$$L = \int_0^1 \sqrt{1 + (y\sqrt{y^2+2})^2} , dy = \int_0^1 \sqrt{1 + y^2(y^2+2)} , dy$$
$$= \int_0^1 \sqrt{1 + y^4 + 2y^2} , dy = \int_0^1 \sqrt{(y^2+1)^2} , dy = \int_0^1 (y^2+1) , dy$$
$$L = [\frac{y^3}{3} + y]_0^1 = 1/3 + 1 = 4/3$$

# Example 24: Polar Area - Inner Loop of Limaçon (BC Level 80)

Find the area of the inner loop of $r = 1 - 2 \sin \theta$.
Solution:
The inner loop exists when $r < 0$. $1 - 2 \sin \theta = 0 \implies \sin \theta = 1/2$.
$\theta = \pi/6$ and $5\pi/6$.
$$\text{Area} = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (1 - 2\sin \theta)^2 , d\theta = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (1 - 4\sin \theta + 4\sin^2 \theta) , d\theta$$
$$= \frac{1}{2} \int_{\pi/6}^{5\pi/6} (1 - 4\sin \theta + 2 - 2\cos 2\theta) , d\theta = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (3 - 4\sin \theta - 2\cos 2\theta) , d\theta$$
$$= \frac{1}{2} [3\theta + 4\cos \theta - \sin 2\theta]_{\pi/6}^{5\pi/6} = \pi - \frac{3\sqrt{3}}{2}$$

# Example 25: Average Value of a Trig Function (Level 30)

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

# Example 26: Volume - Isosceles Right Triangles (Level 50)

Base is the region between $y=x^2$ and $y=1$. Cross sections perpendicular to the $y$-axis are isosceles right triangles with the hypotenuse in the base.
Solution:
Bounds for $y$: $0$ to $1$. Base length $s$ of the triangle is the width of the region at $y$.
$x = \pm \sqrt{y} \implies s = 2\sqrt{y}$.
Area of isosceles right triangle with hypotenuse $s$: $A(y) = \frac{1}{4}s^2$.
$$A(y) = \frac{1}{4}(2\sqrt{y})^2 = \frac{1}{4}(4y) = y$$
$$V = \int_0^1 y , dy = [\frac{y^2}{2}]_0^1 = 1/2$$

# Example 27: Position from Acceleration (Level 40)

$a(t) = \cos t, v(0) = 5, s(0) = 0$. Find $s(\pi)$.
Solution:
$v(t) = \int a(t) , dt = \sin t + C$. Since $v(0)=5$, $0+C=5 \implies v(t) = \sin t + 5$.
$s(t) = \int v(t) , dt = -\cos t + 5t + C_2$. Since $s(0)=0$, $-1 + 0 + C_2 = 0 \implies C_2 = 1$.
$s(t) = -\cos t + 5t + 1$.
$s(\pi) = -(-1) + 5\pi + 1 = 5\pi + 2$.

# Example 28: Total Net Change in Population (Level 30)

Population growth rate is $P'(t) = 400 + 60\sqrt{t}$ people/year. Find the increase in population from year 4 to year 9.
Solution:
$$\Delta P = \int_4^9 (400 + 60t^{1/2}) , dt = [400t + 40t^{3/2}]_4^9$$
$$= (3600 + 40 \cdot 27) - (1600 + 40 \cdot 8) = (3600 + 1080) - (1600 + 320) = 4680 - 1920 = 2760$$

# Example 29: Generalizing Volume (Level 90)

A solid has a base bounded by $f(x)$ and $g(x)$. If the cross-sections are any shape with area $k[s(x)]^n$, show the volume formula.
Solution:
$s(x) = f(x) - g(x)$.
$V = \int_a^b k[f(x) - g(x)]^n , dx$.
For squares, $k=1, n=2$. For semicircles, $k=\pi/8, n=2$. This demonstrates that the geometry of the cross-section is simply a constant multiplier if the area is proportional to the square of the base.

# Example 30: Integrating a Rate (Level 45)

A pipe leaks oil at $R(t) = \frac{10}{t+1}$ gal/hr. How much oil leaks during the first 10 hours?
Solution:
$$\int_0^{10} \frac{10}{t+1} , dt = 10 [\ln|t+1|]_0^{10} = 10(\ln 11 - \ln 1) = 10 \ln 11 \approx 23.98 \text{ gal.}$$

# Example 31: Advanced Washer Method (Level 85)

Region bounded by $y = \ln x$, $y=0$, $x=e$. Rotate about the line $x=-2$.
Solution:
Since the axis is vertical, integrate $dy$. $y = \ln x \implies x = e^y$.
Bounds for $y$: $0$ to $1$.
$R_{out} = \text{distance from } x=-2 \text{ to } x=e \implies e - (-2) = e+2$.
$r_{in} = \text{distance from } x=-2 \text{ to } x=e^y \implies e^y - (-2) = e^y + 2$.
$$V = \pi \int_0^1 [(e+2)^2 - (e^y+2)^2] , dy = \pi \int_0^1 (e^2 + 4e + 4 - (e^{2y} + 4e^y + 4)) , dy$$
$$= \pi \int_0^1 (e^2 + 4e - e^{2y} - 4e^y) , dy = \pi [ (e^2+4e)y - \frac{1}{2}e^{2y} - 4e^y ]_0^1$$
$$= \pi [(e^2+4e - \frac{1}{2}e^2 - 4e) - (0 - 1/2 - 4)] = \pi [ \frac{1}{2}e^2 + 4.5 ]$$

# Example 32: Area of a Rose Curve (BC Level 70)

Find the area of one petal of $r = \cos 3\theta$.
Solution:
One petal is traced as $r$ goes from $0$ to max and back to $0$.
$\cos 3\theta = 0 \implies 3\theta = -\pi/2, \pi/2 \implies \theta = -\pi/6, \pi/6$.
$$\text{Area} = \frac{1}{2} \int_{-\pi/6}^{\pi/6} \cos^2(3\theta) , d\theta = \frac{1}{2} \int_{-\pi/6}^{\pi/6} \frac{1 + \cos 6\theta}{2} , d\theta$$
$$= \frac{1}{4} [\theta + \frac{1}{6}\sin 6\theta]_{-\pi/6}^{\pi/6} = \frac{1}{4} [(\pi/6 + 0) - (-\pi/6 + 0)] = \frac{\pi}{12}$$

# Example 33: Function vs Inverse Area (Level 75)

Show that the area under $f(x)$ from $a$ to $b$ plus the area under $f^{-1}(x)$ from $f(a)$ to $f(b)$ is $b \cdot f(b) - a \cdot f(a)$.
Solution:
Let $y = f(x)$, so $x = f^{-1}(y)$ and $dy = f'(x)dx$.
Area 2 $= \int_{f(a)}^{f(b)} f^{-1}(y) , dy = \int_a^b x f'(x) , dx$.
Using integration by parts: $u=x, dv=f'(x)dx \implies du=dx, v=f(x)$.
$\int_a^b x f'(x) , dx = [x f(x)]_a^b - \int_a^b f(x) , dx$.
Therefore, $\int_a^b f(x) , dx + \text{Area 2} = b f(b) - a f(a)$.

# Example 34: Surface Area of Revolution (Optional/Advanced Context Level 95)

Find the surface area of the solid generated by rotating $y = x^3$ from $0 \le x \le 1$ about the $x$-axis.
Solution:
Formula: $S = \int 2\pi y , ds = \int_0^1 2\pi x^3 \sqrt{1 + (3x^2)^2} , dx$.
$$S = 2\pi \int_0^1 x^3 \sqrt{1 + 9x^4} , dx$$
Let $u = 1+9x^4, du=36x^3 dx$.
$$S = 2\pi \cdot \frac{1}{36} \int_1^{10} u^{1/2} , du = \frac{\pi}{18} [\frac{2}{3}u^{3/2}]_1^{10} = \frac{\pi}{27}(10\sqrt{10} - 1)$$

# Example 35: The Volume of a Sphere (Level 100)

Derive the volume of a sphere of radius $R$ using calculus.
Solution:
Rotate the semicircle $y = \sqrt{R^2 - x^2}$ from $x = -R$ to $x = R$ about the $x$-axis.
Using the Disk Method:
$$V = \pi \int_{-R}^R (\sqrt{R^2 - x^2})^2 , dx = \pi \int_{-R}^R (R^2 - x^2) , dx$$
$$V = \pi [R^2x - \frac{x^3}{3}]_{-R}^R = \pi [(R^3 - R^3/3) - (-R^3 + R^3/3)]$$
$$V = \pi [2R^3/3 - (-2R^3/3)] = \frac{4}{3}\pi R^3$$