Unit 8: Applications of Integration

Average Value of a Function

Formula, Interpretation, and How to Use It

• The average value of a continuous function on an interval is \(f_{\text{avg}}=\dfrac{1}{b-a}\int_a^b f(x)\,dx\). This formula comes from “total accumulated height” divided by “length of the interval,” exactly mirroring average speed as total distance over total time. It turns an area (the definite integral) into a representative constant height that would produce the same area if drawn as a horizontal strip.
• To compute \(f_{\text{avg}}\), first evaluate the definite integral with the Fundamental Theorem of Calculus, then divide by \((b-a)\). If the integral is hard symbolically, approximate it using geometry (triangles, rectangles, trapezoids) or numerical rules like trapezoidal. Always attach correct units, which are the same as the function’s units (not “units squared”).
• Worked mini-example: For \(f(x)=x^2\) on \([0,3]\), \(\int_0^3 x^2\,dx=\left[\tfrac{x^3}{3}\right]_0^3=9\), so \(f_{\text{avg}}=\tfrac{1}{3}\cdot 9=3\). The line \(y=3\) has the same area over \([0,3]\) as the region under \(y=x^2\), which is the precise geometric interpretation of average value.
• Reasoning: The integral aggregates signed “heights,” so if parts of \(f\) are negative, the average can be below zero. This is why the average value depends on signed area, unlike an average of absolute heights; knowing this prevents misinterpreting contexts where negative contributions matter (for example, velocity to the left).
• Common pitfalls: confusing \(f_{\text{avg}}\) with \(\tfrac{f(a)+f(b)}{2}\), which is only valid for linear functions; forgetting to divide by \((b-a)\), which yields total accumulation not an average; and losing units. A good check is to ensure \(f_{\text{avg}}\) has the same units as \(f\).

Finding Points Where \(f(x)=f_{\text{avg}}\) (MVT for Integrals)

• The Mean Value Theorem for Integrals guarantees a point \(c\in[a,b]\) with \(f(c)=f_{\text{avg}}\) when \(f\) is continuous. This provides existence, not a formula for \(c\), so in practice you compute \(f_{\text{avg}}\) first, then solve \(f(x)=f_{\text{avg}}\). Having a concrete value for the average turns the existence theorem into a solvable equation.
• Worked mini-example: For \(f(x)=\sin x\) on \([0,\pi]\), \(f_{\text{avg}}=\dfrac{1}{\pi}\int_0^\pi \sin x\,dx=\dfrac{2}{\pi}\). Solve \(\sin x=\tfrac{2}{\pi}\) to find the times when instantaneous height equals the average height, usually two symmetric solutions inside the interval.
• Reasoning: Because \(\int_a^b (f(x)-f_{\text{avg}})\,dx=0\), regions where \(f\) is above the average must balance regions where it is below. This balance view helps you estimate where solutions might live even before solving algebraically, especially with graphs or tables.
• Connections: This topic ties to Unit 6’s accumulation functions \(F(x)=\int_a^x f(t)\,dt\). Since \(F'(x)=f(x)\), horizontal tangents of the line \(y=f_{\text{avg}}\) correspond to equal-area reasoning for \(F\), linking averages to area balance and derivative signs.
• AP tips and pitfalls: Put calculators in radian mode for trig; watch domain restrictions when solving \(f(x)=\) constant; and if \(f\) is not one-to-one, expect multiple \(c\)-values. When a table is provided, approximate the integral first (e.g., trapezoid) and then divide by \((b-a)\).

Modeling Particle Motion

Displacement vs. Total Distance (How to Compute Step by Step)

• Displacement on \([t_1,t_2]\) is the signed integral \(\displaystyle \int_{t_1}^{t_2} v(t)\,dt\), which adds positive motion and subtracts motion in the opposite direction. This directly uses Unit 6’s “integral of rate equals net change” principle with velocity as the rate. Always interpret sign: negative results mean net movement in the negative direction.
• Total distance is \(\displaystyle \int_{t_1}^{t_2}\!|v(t)|\,dt\), which requires splitting at the zeros of \(v\). The workflow is: solve \(v(t)=0\), partition the interval, remove absolute value by assigning the correct sign on each subinterval, and integrate piecewise. This prevents cancellation that would hide actual path length.
• Worked mini-example: \(v(t)=t^2-4t+3=(t-1)(t-3)\) on \([0,4]\). Displacement is \(\int_0^4 v(t)\,dt\) which evaluates to \( \left[\tfrac{t^3}{3}-2t^2+3t\right]_0^4= \tfrac{64}{3}-32+12= \tfrac{4}{3}\). Total distance splits at \(t=1,3\); compute each piece’s magnitude and sum to avoid cancellations.
• Reasoning: The sign of \(v\) tells direction, and the magnitude \(|v|\) controls “speed” of accumulation in distance. Checking signs first provides a roadmap for both conceptual interpretation and correct computation, which is especially important on free-response.
• Common pitfalls: forgetting to split at zeros when integrating \(|v|\), losing units (distance should be in length units), and mixing up displacement with total distance. A quick units check and sign chart typically catch these errors.

From Acceleration to Velocity and Position (Using Initial Conditions)

• When \(a(t)\) is given, integrate once to recover velocity: \(v(t)=v(t_0)+\int_{t_0}^{t} a(s)\,ds\). Then integrate velocity to recover position: \(x(t)=x(t_0)+\int_{t_0}^{t} v(s)\,ds\). These are direct applications of FTC and net change, with initial conditions anchoring the constants.
• Worked mini-example: \(a(t)=6t\), \(v(0)=4\), \(x(0)=3\). First, \(v(t)=4+\int_0^t 6s\,ds=4+3t^2\). Next, \(x(t)=3+\int_0^t (4+3s^2)\,ds=3+4t+t^3\). Each step includes evaluation at bounds and is validated by differentiating back to check consistency.
• Reasoning: Each integral accumulates the effect of a rate, so units progress from \(\text{m/s}^2\) to \(\text{m/s}\) to \(\text{m}\). Integrating with bounds (rather than “+C” first) keeps constants organized and reduces algebraic mistakes when multiple initial values are given.
• Connections: Sign analysis of \(v\) and zeros of \(v\) link to displacement and total distance, while concavity of \(x\) is controlled by acceleration \(a=x''\). This mirrors Unit 5 derivative tests, now interpreted within motion.
• Common pitfalls: dropping constants, misreading initial conditions, and evaluating integrals without matching the correct bounds. A clean computation ladder (acceleration → velocity → position) and consistent bounds avoid these errors.

Solving Accumulation Problems

Net Change Theorem in Context (How-To Template)

• Template: identify the rate \(R(t)\) of the quantity of interest and the time window \([a,b]\), then compute total change as \(\displaystyle \int_a^b R(t)\,dt\). Add this to the initial amount to get the final amount: \(\text{final}=\text{initial}+\int_a^b R(t)\,dt\). This single sentence converts word problems into math consistently across physics, biology, and economics.
• Worked mini-example (inflow/outflow): If a tank starts with \(V_0\) liters and fills at \(r_{\text{in}}(t)\) while draining at \(r_{\text{out}}(t)\), the amount at time \(b\) is \(V(b)=V_0+\int_a^b\!\big(r_{\text{in}}(t)-r_{\text{out}}(t)\big)\,dt\). The sign convention “in minus out” ensures correct direction for accumulation.
• Reasoning: Integrals aggregate rate “instants” over time, so the units of the integral are the quantity’s units (rate × time). This dimensional check is your fastest correctness filter and should be stated when justifying answers on FRQs.
• Connections: This is the same logic as motion with velocity and distance, only with a different context-specific rate. It also connects to Unit 7 differential equations, where solving a separable DE essentially integrates a rate law to recover the quantity.
• Pitfalls: forgetting signs (especially when rates can be negative), mixing initial values from different times, and averaging rates incorrectly instead of integrating. If only a table is provided, use trapezoidal rule to approximate the integral and clearly state over/underestimation based on concavity.

Piecewise Rates, Absolute Values, and Units/Reasonableness Checks

• For piecewise rates, split the integral at the breakpoints and integrate each piece with its own formula before summing. This mirrors how you handled piecewise velocity for total distance and ensures you respect the definition on each subinterval. Clean bounds and labels are crucial for partial credit.
• When the question asks for “total amount accumulated” independent of sign (e.g., total rainfall regardless of evaporation), integrate the absolute value of the signed rate. You must first identify where the rate changes sign and treat each segment accordingly.
• Worked mini-example (table): Given equally spaced times \(t_0,\ldots,t_n\) and a rate column \(R(t_i)\), approximate \(\int_a^b R(t)\,dt\) with \(T=\tfrac{\Delta t}{2}\big(R_0+2\sum R_i+R_n\big)\). State whether \(T\) is an over- or under-estimate using concavity or monotonicity clues if provided.
• Reasonableness: compare your final value to simple bounds like \((b-a)\cdot \min R\) and \((b-a)\cdot \max R\). If your answer lies outside these, re-check arithmetic or sign choices; this quick bounding argument is an AP-friendly way to justify accuracy.
• Common pitfalls: integrating across discontinuities without splitting, ignoring units in the final answer, and forgetting that “net” and “total” are different. Write “net change = signed accumulation” and “total = absolute accumulation” in your notes to avoid last-minute confusion.

Solving Accumulation Problems

• An accumulation problem involves determining the net amount of a quantity that has built up over a time interval, given its rate of change. If \( F'(x) = f(x) \) represents the rate, then \( F(b) - F(a) = \int_a^b f(x)\,dx \) gives the net change from \( x = a \) to \( x = b \). This allows us to connect instantaneous rates with total changes.
• The initial value of the quantity is critical to finding the total amount present at a later time. The formula becomes \( F(b) = F(a) + \int_a^b f(x)\,dx \), where \( F(a) \) is the starting amount. This is why initial conditions are always explicitly given in AP problems.
• In real-world applications, the rate function \( f(x) \) might represent velocity, growth rate, or flow rate. The integral can then represent displacement, total growth, or total inflow/outflow. In some cases, absolute value is needed if the problem asks for total distance traveled instead of displacement.
• For example, if a tank initially contains 50 liters of water and water flows in at a rate of \( r(t) = 5\sqrt{t} \) liters per minute for \( 0 \leq t \leq 9 \), then the total amount after 9 minutes is \( 50 + \int_0^9 5\sqrt{t} \, dt \), which we can evaluate directly.

Finding the Area Between Curves

• The area between curves measures the total space enclosed vertically between two functions over a specific interval. If \( f(x) \geq g(x) \) on the interval \([a, b]\), the area is given by \( A = \int_a^b [f(x) - g(x)] \, dx \). This formula subtracts the lower curve from the upper curve point-by-point.
• When curves intersect within the interval, the problem must be split into subintervals where the "top" function remains consistent. Intersections are found by solving \( f(x) = g(x) \). This is especially important in AP problems where graphs change which function is on top.
• If the curves are given in terms of \( y \) (horizontal slices), we integrate with respect to \( y \) instead: \( A = \int_{y_{\text{min}}}^{y_{\text{max}}} [x_{\text{right}}(y) - x_{\text{left}}(y)] \, dy \). This approach is required when vertical slices are impractical due to function orientation.
• Graphical interpretation is crucial — sketching the curves ensures the correct order of subtraction and prevents negative area errors. Common mistakes include forgetting to split intervals or integrating the wrong variable.

Determining Volume with Cross-Sections

• If a solid has a known base in the \( xy \)-plane and a known cross-sectional shape perpendicular to an axis, its volume can be computed using \( V = \int_a^b A(x) \, dx \), where \( A(x) \) is the area of the cross-section at position \( x \). The key step is expressing \( A(x) \) in terms of the base dimensions.
• Common cross-sectional shapes include squares, semicircles, equilateral triangles, and rectangles. For example, if the cross-section is a square of side length \( s(x) \), then \( A(x) = [s(x)]^2 \). If it is a semicircle of diameter \( d(x) \), then \( A(x) = \frac{\pi}{8}[d(x)]^2 \).
• The limits of integration \([a, b]\) are determined by the base region’s projection on the axis of integration. If integrating with respect to \( y \), use \( V = \int_{y_{\min}}^{y_{\max}} A(y) \, dy \).
• Visualizing the solid and its slices is essential. Many AP questions will give the base region explicitly and specify the type of cross-section, requiring you to construct \( A(x) \) or \( A(y) \) from geometric reasoning before integrating.

Finding the Area Between Curves

Concept and Setup

• The area between two curves is found by integrating the difference between the top (or right) function and the bottom (or left) function over a given interval. For vertical slices, use \( \text{Area} = \int_a^b [f(x) - g(x)] \, dx \), where \( f(x) \) is the upper function and \( g(x) \) is the lower. This works because integration accumulates the “strip” heights across the interval.
• For horizontal slices, the formula changes to \( \text{Area} = \int_c^d [f(y) - g(y)] \, dy \), where \( f(y) \) is the right function and \( g(y) \) is the left. Choosing horizontal vs. vertical slicing depends on whether the functions are easier to express in terms of \( x \) or \( y \).
• Always sketch the functions first to identify which curve is on top (or right) across the interval. The top/bottom relationship can change within the interval, requiring you to split the integral at intersection points.

Example

• Example: Find the area between \( y = x^2 \) and \( y = x + 2 \). First, find intersection points by solving \( x^2 = x + 2 \), giving \( x = -1 \) and \( x = 2 \). Here, \( y = x + 2 \) is on top between these points.
• Set up the integral: \( \int_{-1}^2 \big[(x + 2) - (x^2)\big] dx \). Integrate to get \( \left[ \frac{1}{2}x^2 + 2x - \frac{1}{3}x^3 \right]_{-1}^2 \). Evaluate: At \( x = 2 \), value = \( 2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} \). At \( x = -1 \), value = \( 0.5 - 2 + \frac{1}{3} = -1.5 + \frac{1}{3} \).
• Subtract to find total area = \( \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \) units².

Common Mistakes

• Mixing up top/bottom curves. Always check with a test point.
• Forgetting to split the integral if curves cross within the bounds.
• Not switching to horizontal slices when vertical slices lead to two separate functions for the top curve.

Determining Volume with Cross-Sections, the Disc Method, and the Washer Method

Disc Method

• When a region is revolved around an axis and forms a solid without a hole, use the disc method. The volume is given by \( V = \pi \int_a^b [R(x)]^2 dx \), where \( R(x) \) is the distance from the axis of rotation to the outer curve. This works because each “disc” has area \( \pi R^2 \) and thickness \( dx \).
• If revolving around the y-axis, switch to \( V = \pi \int_c^d [R(y)]^2 dy \) where \( R(y) \) is horizontal distance from the axis to the curve.

Washer Method

• When a region is revolved and forms a hollow solid, use the washer method. The volume is \( V = \pi \int_a^b \big([R_{\text{outer}}(x)]^2 - [R_{\text{inner}}(x)]^2\big) dx \), subtracting the “hole” from the outer disc.
• Always square both the outer and inner radii before subtracting. Forgetting to square before subtraction is a common mistake that changes the result entirely.

Cross-Section Method

• If the cross-section shape is known (square, triangle, semicircle, etc.), the volume is \( V = \int_a^b A(x) \, dx \) where \( A(x) \) is the area of the cross-section at position \( x \). For example, a square cross-section with base \( b(x) \) has area \( b(x)^2 \).
• Example: For a semicircular cross-section, \( A(x) = \frac{1}{2} \pi \left(\frac{b(x)}{2}\right)^2 \).

Connections

• Relates to Unit 6 (Definite integrals) since volume is just the accumulation of infinitesimally thin slices.
• Requires skills from Unit 2 (Functions) to set up \( R(x) \) correctly from a graph or equation.

Determining the Length of a Planar Curve Using a Definite Integral

Formula

• 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 pieces of the curve.
• If the curve is given as \( x = g(y) \), the formula becomes: \[ L = \int_c^d \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \]
• For parametric equations \( x(t), y(t) \), use: \[ L = \int_{\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]

Example

• Example: Find the length of \( y = \frac{x^{3/2}}{3} \) from \( x = 0 \) to \( x = 4 \). First, \( \frac{dy}{dx} = \frac{\sqrt{x}}{2} \). Then \( \left(\frac{dy}{dx}\right)^2 = \frac{x}{4} \).
• Arc length formula: \( L = \int_0^4 \sqrt{1 + \frac{x}{4}} \, dx \). Factor: \( L = \int_0^4 \sqrt{\frac{x + 4}{4}} \, dx = \frac{1}{2} \int_0^4 \sqrt{x + 4} \, dx \).
• Let \( u = x + 4 \), \( du = dx \), bounds: \( u: 4 \to 8 \). Integral: \( L = \frac{1}{2} \cdot \frac{2}{3} [u^{3/2}]_4^8 = \frac{1}{3} \left( 8^{3/2} - 4^{3/2} \right) = \frac{1}{3} ( 16\sqrt{2} - 8 ) \) units.