Unit 8: Applications of Integration

Finding the Average Value of a Function

• The average value of a continuous function \( f(x) \) on an interval \([a, b]\) is given by \( f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx \). This formula takes the total accumulated value of the function over the interval and divides it evenly, just like finding the mean of a set of numbers. It’s important to remember that the average value is a horizontal “height” that would produce the same area if the function were constant.
• To calculate it, you must first set up the correct bounds \([a, b]\), integrate \(f(x)\), and then divide the result by \(b-a\). A common mistake is to forget the \( \frac{1}{b-a} \) factor, which means you’d just be finding total accumulation instead of an average.
• Average value problems often appear in physical contexts, like finding the average velocity over a time period or average temperature over a day. In these cases, be sure your \(f(x)\) matches the quantity in question and that you interpret units properly — for example, average velocity will be in meters per second if \(f(t)\) is velocity in meters per second.
• This concept connects directly to the Mean Value Theorem for Integrals, which guarantees that there is at least one \(c\) in \([a, b]\) where \( f(c) = f_{\text{avg}} \). Understanding this connection helps with interpreting graphical problems where you visually identify the height that “balances” the function’s area.
• When solving on the AP exam, always check whether the question is asking for the average value of the function itself or the average rate of change — these are different and use different formulas. The former uses integration, the latter uses slope formulas from Unit 2.

Modeling Particle Motion

• Particle motion problems in integration focus on using position, velocity, and acceleration functions to describe movement along a line. If velocity \(v(t)\) is given, the net change in position over \([a, b]\) is \( \int_a^b v(t) \, dt \). This integral can be positive or negative, depending on direction, so it represents displacement, not total distance traveled.
• To find total distance traveled, you integrate the speed \( |v(t)| \) instead, which requires determining when velocity changes sign. This usually means solving \( v(t) = 0 \) for \(t\) and splitting the integral into intervals where velocity is positive or negative, then adding the absolute values of each displacement.
• If acceleration \(a(t)\) is given, you integrate it to find velocity, remembering to include the constant from the initial velocity condition. Similarly, if velocity is given, integrating gives position. This connection is exactly the reverse of differentiation from Units 2 and 4.
• On the AP exam, motion problems often include interpretation questions — for example, “When is the particle moving to the right?” (where \(v(t) > 0\)) or “When is the particle speeding up?” (where \(v(t)\) and \(a(t)\) have the same sign). Being fluent in these conditions is essential for quick problem solving.
• Graphical versions may give you a velocity-time graph instead of an equation. In these cases, displacement is the signed area under the curve, and total distance is the sum of absolute areas. Pay close attention to areas below the axis, as they represent motion in the opposite direction.

Solving Accumulation Problems

• Accumulation problems ask how much of a quantity builds up over time when you know its rate of change. The general formula is \( Q(t) = Q(a) + \int_a^t R(x) \, dx \), where \(Q(a)\) is the initial amount and \(R(x)\) is the rate. This formula works for any quantity: water in a tank, money in an account, or bacteria in a culture.
• The key to setting these up correctly is identifying the correct initial value and ensuring units match between the rate function and the quantity being accumulated. If the rate is given in liters per minute, integrating over minutes gives liters, which you then add to the initial liters in the tank.
• Some AP problems include rates that switch between positive (adding) and negative (removing) values, which means the quantity may increase and then decrease. Always check the sign of the rate before interpreting results, and split the integral if necessary.
• Graph-based accumulation problems require interpreting the area under a rate curve. If the rate is above the x-axis, the quantity is increasing; if below, the quantity is decreasing. Calculating exact area may involve geometry formulas or breaking curves into segments.
• This topic is one of the most common free-response question types on the AP exam because it connects conceptual understanding, integration skills, and real-world application. Being fluent in switching between rate and total quantity is essential for success.

Area Between Curves

• To find the area between two curves \( y_{\text{top}} \) and \( y_{\text{bottom}} \) on an interval \([a, b]\), use \( \text{Area} = \int_a^b [y_{\text{top}} - y_{\text{bottom}}] \, dx \). First, sketch the graphs or visualize them to determine which function is on top for the given interval. If the top and bottom functions switch positions, split the integral at the intersection point(s).
• If the curves are given as functions of \(x\), integrate with respect to \(x\). If they are functions of \(y\), integrate with respect to \(y\) using \( x_{\text{right}} - x_{\text{left}} \) inside the integral. Choosing the correct orientation can make the integral easier.
• When curves intersect, set them equal to each other to find the intersection points — these are the limits of integration. Forgetting this step or mixing up which curve is on top is the most common mistake on the AP exam for this topic.
• For complicated functions, technology (graphing calculators) can help verify which function is on top and calculate intersection points accurately. However, the setup of the integral must be correct; the calculator only evaluates what you give it.
• This topic connects directly to geometry: the area between curves is simply the sum of many vertical (or horizontal) slices, each with height equal to the difference between the curves. Understanding this geometric meaning helps with more advanced applications like volumes.

Volumes Using Cross Sections

• If a solid has a known base region in the plane and the cross section perpendicular to an axis is a known shape, the volume is \( V = \int_a^b A(x) \, dx \), where \( A(x) \) is the area of the cross section at position \(x\). The base is usually described by curves or boundaries in the plane, and the cross section shape tells you the formula for \(A(x)\).
• Common cross sections include squares (\(s^2\)), semicircles (\(\frac{1}{2} \pi r^2\)), equilateral triangles (\(\frac{\sqrt{3}}{4} s^2\)), and rectangles (\( \text{length} \times \text{width} \)). The dimension \(s\) or \(r\) is determined by the width of the base at a given \(x\) or \(y\).
• Step-by-step: (1) Draw the base and mark a slice. (2) Express the slice’s width in terms of \(x\) or \(y\). (3) Use the shape’s area formula for \(A(x)\). (4) Integrate over the correct bounds to get volume. Forgetting to square side lengths when needed is a common AP exam mistake.
• When the base is between curves, the width of the slice is \((\text{right function} - \text{left function})\) if slicing vertically, or \((\text{top function} - \text{bottom function})\) if slicing horizontally. This width then goes into the area formula for the cross section.
• Understanding this method makes later topics like volumes of revolution easier, since both are just applications of slicing and adding infinitely many thin pieces.

Volumes Using the Disk and Washer Methods

• The disk method finds the volume of a solid of revolution when the cross sections perpendicular to the axis of revolution are solid circles. If rotating around the \(x\)-axis, the formula is \( V = \pi \int_a^b [R(x)]^2 \, dx \), where \(R(x)\) is the distance from the axis to the curve.
• The washer method is used when the solid has a hole in the middle. The formula becomes \( V = \pi \int_a^b \left( [R_{\text{outer}}(x)]^2 - [R_{\text{inner}}(x)]^2 \right) dx \). Here, you subtract the inner radius’s area from the outer radius’s area before integrating.
• Step-by-step: (1) Draw the region and axis of rotation. (2) Identify inner and outer radii. (3) Square them because volume is based on area of circles. (4) Integrate along the axis of revolution. Forgetting to square the radii or mixing inner/outer radii is a high-frequency error on AP questions.
• If revolving around the \(y\)-axis, express functions as \(x\) in terms of \(y\) and integrate with respect to \(y\). The setup changes, but the geometric principle — summing disks or washers — is the same.
• These methods rely on symmetry and the geometry of circles, making them powerful for solids with curved boundaries. Always visualize the solid before deciding between disk or washer methods.

Volumes Using the Shell Method

• The shell method is used to find the volume of a solid of revolution by summing cylindrical shells rather than disks or washers. The formula is \( V = 2\pi \int_a^b [\text{radius}] \cdot [\text{height}] \, dx \) if integrating along the \(x\)-axis, or \( V = 2\pi \int_a^b [\text{radius}] \cdot [\text{height}] \, dy \) if integrating along the \(y\)-axis.
• The radius is the distance from the axis of rotation to the slice, and the height is the length of the function in the direction perpendicular to the axis. Always visualize the shell as a hollow tube whose thickness is \(dx\) or \(dy\).
• Step-by-step: (1) Draw the region and axis of rotation. (2) Identify which way you want to slice — vertical slices work well for rotation around the \(y\)-axis, horizontal slices for rotation around the \(x\)-axis. (3) Write expressions for radius and height. (4) Plug into the formula and integrate.
• The shell method is often easier than the disk/washer method if the axis of rotation is not the main coordinate axis, or if solving for one variable in terms of the other is messy. Choosing the right method can save time and reduce errors on AP questions.
• Remember: If the axis of rotation is offset (e.g., \(y = -2\)), adjust the radius accordingly. Forgetting to adjust the radius for vertical/horizontal shifts is a common mistake.

Arc Length and Surface Area of Revolution

• The arc length of a curve \(y = f(x)\) from \(x = a\) to \(x = b\) is given by \( L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \). This formula comes from the Pythagorean theorem applied to infinitesimally small segments of the curve.
• For curves given as \(x = g(y)\), the formula becomes \( L = \int_c^d \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \). Choosing the right variable to integrate with respect to can simplify calculations if one derivative is easier than the other.
• Surface area of a solid of revolution is found by multiplying the arc length element by the circumference of the rotation: \( S = 2\pi \int_a^b [\text{radius}] \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \) for rotation around the \(x\)-axis, or similar with \(dy\) for \(y\)-axis rotation.
• Step-by-step: (1) Differentiate your function. (2) Square the derivative and add 1. (3) Take the square root and integrate over the correct bounds. (4) For surface area, include the radius factor. This is a direct extension of the arc length formula, just with the circular motion included.
• Arc length problems often require more algebraic manipulation and trigonometric substitution than basic area/volume problems. Be comfortable simplifying square roots and recognizing when an integral can be evaluated exactly or needs numerical methods.

Practice Problem 1 — Shell Method and Area Between Curves

Question: The region bounded by \( y = \sqrt{x} \), \( y = 0 \), and \( x = 4 \) is revolved about the line \(x = -1\). Use the shell method to find the volume of the solid generated.

Solution:

• Step 1: Identify the shape of slices. Because we are rotating around a vertical line \(x = -1\) and the given boundaries are in terms of \(x\), we will use vertical slices, which means our shells will have radius \(r = x - (-1) = x + 1\) and height \(h = \sqrt{x}\).
• Step 2: Write the formula for volume using the shell method: \[ V = 2\pi \int_0^4 (\text{radius}) \cdot (\text{height}) \, dx \] Substituting \(r = x+1\) and \(h = \sqrt{x}\): \[ V = 2\pi \int_0^4 (x+1)\sqrt{x} \, dx \]
• Step 3: Expand and integrate: \[ (x+1)\sqrt{x} = x^{3/2} + x^{1/2} \] \[ V = 2\pi \left[ \frac{2}{5}x^{5/2} + \frac{2}{3}x^{3/2} \right]_{0}^{4} \]
• Step 4: Evaluate at bounds: \[ x^{5/2} \text{ at } x=4 = (4^{5/2}) = ( \sqrt{4} )^5 = 2^5 = 32 \] \[ x^{3/2} \text{ at } x=4 = (\sqrt{4})^3 = 2^3 = 8 \] \[ V = 2\pi \left[ \frac{2}{5}(32) + \frac{2}{3}(8) \right] \]
• Step 5: Simplify: \[ V = 2\pi \left[ \frac{64}{5} + \frac{16}{3} \right] \] Common denominator = 15: \[ V = 2\pi \left[ \frac{192}{15} + \frac{80}{15} \right] = 2\pi \left[ \frac{272}{15} \right] = \frac{544\pi}{15} \ \text{units}^3 \]

Practice Problem 2 — Arc Length

Question: Find the length of the curve \( y = \frac{1}{3}x^{3/2} \) from \(x = 0\) to \(x = 4\).

Solution:

• Step 1: Recall the arc length formula: \[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]
• Step 2: Differentiate \(y\): \[ y = \frac{1}{3}x^{3/2} \quad \Rightarrow \quad \frac{dy}{dx} = \frac{1}{3} \cdot \frac{3}{2} x^{1/2} = \frac{1}{2} \sqrt{x} \]
• Step 3: Square the derivative and add 1: \[ \left( \frac{dy}{dx} \right)^2 = \frac{x}{4} \] \[ 1 + \frac{x}{4} = \frac{4 + x}{4} \]
• Step 4: Set up the integral: \[ L = \int_0^4 \sqrt{\frac{4 + x}{4}} \, dx = \frac{1}{2} \int_0^4 \sqrt{x + 4} \, dx \]
• Step 5: Substitute \(u = x + 4\), \(du = dx\), bounds change: when \(x=0, u=4\); when \(x=4, u=8\). \[ L = \frac{1}{2} \int_4^8 \sqrt{u} \, du \] \[ L = \frac{1}{2} \cdot \left[ \frac{2}{3} u^{3/2} \right]_4^8 \]
• Step 6: Evaluate: \[ L = \frac{1}{3} \left[ (8)^{3/2} - (4)^{3/2} \right] = \frac{1}{3} \left[ (\sqrt{8})^3 - (\sqrt{4})^3 \right] \] \[ = \frac{1}{3} \left[ (2\sqrt{2})^3 - 8 \right] = \frac{1}{3} \left[ 16\sqrt{2} - 8 \right] \] Final answer: \[ L = \frac{16\sqrt{2} - 8}{3} \ \text{units} \]

Common Misconceptions

• Students often confuse when to use the disk/washer method versus the shell method. The choice depends on the axis of rotation and the orientation of slices. A good check is to sketch the region and rotation axis—if your slices would require solving for \(y\) when you prefer \(x\) (or vice versa), you may need the other method.
• For arc length problems, students sometimes forget that the derivative \(\frac{dy}{dx}\) must be squared before adding 1 inside the square root. Skipping the squaring step or incorrectly simplifying \(\sqrt{1 + (\frac{dy}{dx})^2}\) can lead to incorrect integrals. Always write the inside of the square root carefully before simplifying.
• When finding areas between curves, students sometimes mix up which function is “top” and which is “bottom” for the interval. The subtraction must always be \( \text{top function} - \text{bottom function} \) (or right minus left in horizontal slices). Using the wrong order can result in negative areas or a completely wrong answer.
• In volume problems, students often forget that all radii and heights must be nonnegative in the formulas. If the region crosses the axis of rotation, you might need to split the integral into separate intervals or take absolute values before integrating.
• In numerical integration (Midpoint, Trapezoidal), students sometimes incorrectly set \(\Delta x\) as the number of subintervals instead of \(\frac{b-a}{n}\). Miscalculating \(\Delta x\) affects all term values and results in large numerical errors. Always find \(\Delta x\) first before plugging into midpoint or trapezoidal formulas.
• For symmetry shortcuts in integrals, students sometimes apply even/odd symmetry rules without checking the function’s domain or whether the bounds are symmetric about zero. Using symmetry incorrectly can cause you to completely miss contributions to the area or volume.