Unit 6: Integration and Accumulation of Change

Students will learn how to apply limits to define definite integrals and how the Fundamental Theorem connects to integration and differentiation.

Accumulation Functions and Definite Integrals

Understanding Accumulation Functions

An accumulation function is defined as \( F(x) = \int_a^x f(t) \, dt \), where \(a\) is the starting point and \(x\) is the variable upper bound. This function represents the net signed area under the curve of \(f(t)\) from \(t=a\) to \(t=x\). It’s important to realize this is not just a number but a new function whose output depends on \(x\).
To evaluate an accumulation function at a specific \(x\), substitute that value into the upper bound and compute the definite integral. For example, if \(F(x) = \int_2^x (t^2 - 1) \, dt\), finding \(F(5)\) means evaluating the definite integral from 2 to 5 of \(t^2 - 1\).
When interpreting \(F(x)\) graphically, positive values of \(f(t)\) add to the accumulation, while negative values subtract from it. This means that \(F(x)\) increases when \(f(t) > 0\) and decreases when \(f(t) < 0\), linking directly to concepts from derivatives.
Accumulation functions connect integrals to rate-of-change problems. For instance, if \(f(t)\) is a velocity function, \(F(x)\) represents the displacement from time \(t=a\) to \(t=x\). This makes them crucial in physical, biological, and economic modeling.
Common mistake: assuming \(F(x)\) always increases with \(x\). Remember, if \(f(t)\) is negative on an interval, the accumulation decreases — a concept directly tied to signed area and direction in motion problems.

Definite Integral Notation and Meaning

The definite integral \(\int_a^b f(x) \, dx\) represents the net signed area between the graph of \(f(x)\) and the \(x\)-axis from \(x=a\) to \(x=b\). Positive regions contribute positively, and negative regions contribute negatively.
To compute a definite integral exactly, find the antiderivative \(F(x)\) of \(f(x)\) and evaluate \(F(b) - F(a)\). This process is guaranteed by the Fundamental Theorem of Calculus (Part 2), which we will explore later.
Definite integrals are tied to units: if \(f(x)\) is in meters/second and \(x\) is in seconds, the integral gives meters. Always track units, especially in applied contexts, to verify the answer’s plausibility.
Graphically, definite integrals can be approximated using Riemann sums, where you sum the areas of rectangles over subintervals. The more subintervals you use, the closer you get to the true value.
Tip: In multiple-choice questions, if the function’s graph is simple, compute areas geometrically (triangles, rectangles, trapezoids) instead of integrating symbolically to save time.

Properties of Integrals and Integration Techniques (Extended)

Properties of Definite Integrals

Linearity: \(\int_a^b [cf(x) + g(x)] \, dx = c\int_a^b f(x) \, dx + \int_a^b g(x) \, dx\). This allows you to split or combine integrals and factor out constants, making complex problems simpler to handle.
Reversing limits: \(\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx\). This is useful when breaking an integral into multiple intervals or correcting an incorrectly ordered set of bounds.
Additivity: \(\int_a^c f(x) \, dx + \int_c^b f(x) \, dx = \int_a^b f(x) \, dx\). This property is essential when integrating piecewise functions or using symmetry to simplify work.
Symmetry: If \(f(x)\) is even, \(\int_{-a}^a f(x) \, dx = 2\int_0^a f(x) \, dx\); if odd, \(\int_{-a}^a f(x) \, dx = 0\). This reduces computation time, especially on the AP exam.
Comparison property: If \(f(x) \leq g(x)\) on \([a,b]\), then \(\int_a^b f(x) \, dx \leq \int_a^b g(x) \, dx\). This can help estimate answers or check if a computed integral is reasonable.

Integration by Substitution

Substitution is the reverse of the chain rule. Let \(u = g(x)\) so that \(du = g'(x)dx\), then rewrite the integral in terms of \(u\) and integrate. This is especially helpful when you see a composite function multiplied by its derivative.
Example: \(\int 2x\cos(x^2) \, dx\), set \(u = x^2\), \(du = 2x\,dx\), giving \(\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C\).
For definite integrals, adjust the bounds to match the new \(u\)-variable instead of converting back to \(x\). This avoids errors and unnecessary extra work.
Substitution is not always necessary in AP Calc AB — sometimes direct integration or geometric interpretation is faster. Always choose the method with the least computation for timed exams.
Common mistake: forgetting to change bounds when doing definite integrals in \(u\)-form, leading to completely incorrect answers.

The Fundamental Theorem of Calculus

Part 1: Derivative of an Accumulation Function

If \(F(x) = \int_a^x f(t) \, dt\), then \(F'(x) = f(x)\). This means the rate of change of the accumulated area is exactly the value of the function at that point. It directly connects differentiation and integration as inverse processes.
If the upper bound is not simply \(x\), apply the chain rule. For example, \(\frac{d}{dx} \int_2^{x^2} \cos(t^3) \, dt = \cos((x^2)^3) \cdot 2x\).
This property allows you to interpret graphs: the slope of \(F(x)\) at any point equals the height of \(f(x)\) at that point. Thus, analyzing \(F\) involves knowing where \(f\) is positive or negative.
Be careful: Part 1 only works if \(f\) is continuous on the interval of integration. If there are discontinuities, the result may not hold without additional considerations.
AP Tip: In free-response, if asked for \(F'(x)\) given an integral, immediately think “replace \(x\) in \(f(t)\) and multiply by the derivative of the bound” — saves precious time.

Evaluating Definite Integrals

If \(F'(x) = f(x)\), then \(\int_a^b f(x) \, dx = F(b) - F(a)\). This turns finding areas into finding antiderivatives, making exact computation much faster than Riemann sums.
Example: \(\int_1^4 (3x^2 - 2) \, dx = [x^3 - 2x]_{1}^{4} = (64 - 8) - (1 - 2) = 56 + 1 = 57\).
FTC Part 2 applies even if \(f\) changes sign — the result is still the net signed area, not the total area. To find total area, integrate \(|f(x)|\) over the same bounds.
When working with piecewise functions, apply FTC separately on each subinterval where \(f\) has a single expression, then add the results.
Common mistake: forgetting parentheses when substituting bounds, which can lead to sign errors in subtraction.

Numerical Integration Methods

Midpoint Rule

The midpoint rule estimates \(\int_a^b f(x) \, dx\) by summing rectangle areas where each rectangle’s height is \(f\) at the midpoint of the subinterval. This generally gives a better estimate than left or right sums when the function is smooth.
To apply: divide \([a,b]\) into \(n\) equal subintervals, find the midpoint of each, evaluate \(f\) there, and multiply by the subinterval width \(\Delta x\). Sum all these products to get the approximation.
Example: \(\int_0^4 (x^2) \, dx\) with \(n=2\). Subintervals: \([0,2]\), \([2,4]\); midpoints: \(x=1, 3\). Heights: \(f(1)=1\), \(f(3)=9\), \(\Delta x = 2\). Estimate = \(2(1) + 2(9) = 2 + 18 = 20\).
The midpoint rule tends to underestimate if the graph is concave up and overestimate if concave down. Recognizing concavity helps predict the direction of error.
Common mistake: using the left or right endpoint instead of the midpoint, which changes the method entirely and gives different accuracy.

Trapezoidal Rule

The trapezoidal rule approximates \(\int_a^b f(x) \, dx\) by summing areas of trapezoids under the curve, formed by connecting consecutive points of \(f(x)\). This method often improves on left/right sums but may be less accurate than midpoint in some cases.
Formula: \(T = \frac{\Delta x}{2} \left[ f(x_0) + 2\sum_{i=1}^{n-1} f(x_i) + f(x_n) \right]\), where \(x_0 = a\) and \(x_n = b\). This weights the endpoints half as much as the interior points.
Example: \(\int_0^4 x^2 \, dx\) with \(n=2\). Points: \(0, 2, 4\); Heights: \(0, 4, 16\); \(\Delta x = 2\). Estimate = \(\frac{2}{2}[0 + 2(4) + 16] = 0 + 8 + 16 = 24\).
Concave up functions cause trapezoidal rule to overestimate, concave down cause underestimation. This is the opposite pattern of the midpoint rule, which can help in error checking.
On the AP Exam, if you have both midpoint and trapezoidal estimates, you can average them for an even better estimate — this is known as Simpson’s Rule (not formally tested in AB, but sometimes useful).

Antiderivatives and Indefinite Integrals

Basic Integration Rules

The antiderivative of \(f(x)\) is a function \(F(x)\) such that \(F'(x) = f(x)\). The general indefinite integral is written \(\int f(x) \, dx = F(x) + C\), where \(C\) is the constant of integration.
Reverse Power Rule: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) for \(n \neq -1\). For \(n = -1\), \(\int x^{-1} dx = \ln|x| + C\).
Example: \(\int (3x^2 - 4x + 7) \, dx = x^3 - 2x^2 + 7x + C\). Each term is integrated separately using the reverse power rule.
Constant multiple rule: \(\int k f(x) \, dx = k \int f(x) \, dx\). This allows factoring out constants before integrating, making calculations simpler.
Common mistake: forgetting \(+C\) in indefinite integrals, which can cost points in FRQs that require the general solution.

Solving for C with Initial Conditions

If given that \(F(x)\) passes through a point \((x_0, y_0)\), substitute \(x_0\) into the antiderivative to solve for \(C\). This yields a particular solution.
Example: \(F'(x) = 6x\), \(F(2) = 5\). Integrate: \(F(x) = 3x^2 + C\). Substituting \(x=2\), \(5 = 3(4) + C \implies C = -7\). So \(F(x) = 3x^2 - 7\).
This step is critical in applied problems, such as finding a position function from a velocity when an initial position is given.
Without an initial condition, you can only find the general form — not the specific one needed for exact predictions.
Tip: Always write down \(+C\) before plugging in the condition; forgetting it can lead to logical errors in solving.

Net Change and Accumulation in Context

Net Change Theorem

The Net Change Theorem states that \(\int_a^b \text{rate} \, dt = \text{(amount at time b)} - \text{(amount at time a)}\). This applies to any scenario where you integrate a rate of change to find total change.
Example: If \(v(t)\) is velocity in m/s, \(\int_0^5 v(t) \, dt\) gives displacement in meters from \(t=0\) to \(t=5\). This is signed displacement — direction matters.
To find total distance traveled, integrate the absolute value of velocity: \(\int_a^b |v(t)| \, dt\). This removes cancellations from direction changes.
Always include units in your answer for applied problems. If \(v(t)\) is in ft/s and \(t\) in seconds, the integral’s units will be feet.
Common mistake: confusing displacement with distance traveled, especially when velocity changes sign.

Piecewise Rates and Applications

When a rate function is defined in pieces, integrate each segment separately over its interval, then sum the results. This ensures you respect the function’s definition in each domain.
Example: \(r(t) = 5\) for \(0 \leq t \leq 3\), and \(r(t) = 10 - t\) for \(3 < t \leq 6\). The total change is \(\int_0^3 5\, dt + \int_3^6 (10-t) \, dt = 15 + \left[10t - \frac{t^2}{2}\right]_3^6 = 15 + (60 - 18) - (30 - 4.5) = 15 + 42 - 25.5 = 31.5\).
In physics, this process is essential for situations like varying acceleration or varying rates of flow in and out of a tank.
Tip: Check if rate changes sign in each piece. If it does, you may need absolute values to get a total magnitude instead of a signed change.
Connection: This method links to properties of integrals (additivity) and graphical analysis — both useful for error checking.

Average Value of a Function

Formula and Concept

The average value of a continuous function \(f\) on \([a,b]\) is given by \(f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx\). This formula essentially takes the total “accumulated height” over the interval and divides it evenly across its length, much like finding average speed.
Example: For \(f(x) = x^2\) on \([0,3]\), the average value is \(\frac{1}{3-0} \int_0^3 x^2 \, dx = \frac{1}{3} \left[ \frac{x^3}{3} \right]_0^3 = \frac{1}{3} \cdot \frac{27}{3} = 3\).
Graphically, \(f_{\text{avg}}\) is the height of a horizontal line that would produce the same area under the curve as \(f\) over \([a,b]\). This can be visualized in shaded-area diagrams.
In applications, average value is used to find average temperatures, average rates, or average concentrations over time or distance, tying directly to real-world modeling.
Common mistake: confusing \(f_{\text{avg}}\) with the arithmetic mean \(\frac{f(a) + f(b)}{2}\). The latter only works for linear functions or when specifically indicated.

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

The Mean Value Theorem for Integrals states that if \(f\) is continuous on \([a,b]\), there exists at least one \(c \in [a,b]\) such that \(f(c) = f_{\text{avg}}\). This is a guarantee, not a guess.
To find such points, set \(f(x) = f_{\text{avg}}\) and solve for \(x\). Sometimes there may be multiple solutions, especially if \(f\) oscillates within the interval.
Example: For \(f(x) = \sin x\) on \([0, \pi]\), \(f_{\text{avg}} = \frac{1}{\pi} \int_0^\pi \sin x \, dx = \frac{1}{\pi} \cdot 2 = \frac{2}{\pi}\). Set \(\sin x = \frac{2}{\pi}\) to find \(x\) values.
Physically, these \(x\)-values represent moments when the instantaneous quantity matches the overall average during the interval.
Tip: When solving in radians, ensure your calculator is in radian mode — mismatched modes can lead to incorrect results on the AP exam.

Motion Applications of Integration

Displacement and Distance

Displacement is given by \(\int_a^b v(t) \, dt\), which produces a signed value based on direction of motion. Negative velocity segments subtract from displacement.
Total distance traveled is \(\int_a^b |v(t)| \, dt\), ensuring all movement contributes positively regardless of direction. This is crucial when velocity changes sign.
Example: If \(v(t) = t^2 - 4\) on \([0,4]\), find where \(v(t)=0\) at \(t=2\), then compute \(\int_0^2 |t^2 - 4| \, dt + \int_2^4 |t^2 - 4| \, dt\) by adjusting signs inside absolute values.
Common mistake: forgetting to split the integral at times when \(v(t) = 0\). This leads to signed displacement instead of total distance.
Tip: Always check for sign changes before integrating velocity — it can completely alter the meaning of the answer.

Position and Velocity from Acceleration

To recover velocity from acceleration, integrate: \(v(t) = \int a(t) \, dt + C\), using an initial velocity to find \(C\). Similarly, position comes from \(s(t) = \int v(t) \, dt + C\).
Example: \(a(t) = 6t\), \(v(0) = 4\). Integrate: \(v(t) = 3t^2 + C\), and using \(v(0)=4\) gives \(C=4\), so \(v(t) = 3t^2 + 4\).
Connecting topics: This process mirrors antiderivatives with initial conditions, as done in indefinite integrals.
Common mistake: mixing up displacement and position — displacement is change in position, not the absolute position itself.
Tip: Keep track of units throughout the process — acceleration in m/s², velocity in m/s, position in meters.

Determining Improper Integrals

Definition and Types

An improper integral is one where the interval of integration is infinite or the integrand becomes unbounded within the interval. These require limits to evaluate.
Type I: Infinite bounds, such as \(\int_1^\infty \frac{1}{x^2} \, dx\). Replace \(\infty\) with a variable limit \(t\) and take \(\lim_{t \to \infty}\) after integrating.
Type II: Unbounded integrand within the interval, such as \(\int_0^1 \frac{1}{\sqrt{x}} \, dx\). Replace the problematic bound with a variable and take a limit as you approach it.
Example: \(\int_1^\infty \frac{1}{x^2} \, dx = \lim_{t\to\infty} \left[ -\frac{1}{x} \right]_1^t = \lim_{t\to\infty} \left( -\frac{1}{t} + 1 \right) = 1\), so it converges to 1.
Common mistake: forgetting to take the limit, leading to treating \(\infty\) as if it were a number — which invalidates the result.

Convergence and Divergence

If the limit process yields a finite number, the integral converges; if it grows without bound or oscillates, it diverges. This classification is critical in determining whether a problem has a meaningful value.
Example of divergence: \(\int_1^\infty \frac{1}{x} \, dx = \lim_{t\to\infty} [\ln|x|]_1^t = \lim_{t\to\infty} \ln t = \infty\), so it diverges.
Improper integrals often appear in applications like probability density functions or physics problems with infinite domains.
On the AP exam, improper integrals are rare but can be tested in conceptual multiple-choice questions involving limits and convergence ideas.
Tip: For quick convergence tests, compare the integrand to a known \(p\)-series \(\frac{1}{x^p}\) — if \(p>1\), it converges; if \(p\leq 1\), it diverges.

Graphical Analysis of Accumulation Functions

From \(f\) to \(F(x)=\int_a^x f(t)\,dt\): Increase, Extrema, and Concavity

To decide where \(F\) increases or decreases, build a quick sign chart for \(f\). If \(f(x) > 0\) on an interval then \(F'(x) = f(x) > 0\) and \(F\) increases there. If \(f(x) < 0\) then \(F\) decreases, so you can read the behavior of \(F\) directly off the vertical position of the graph of \(f\) relative to the axis.
Local maxima and minima of \(F\) occur where \(F'(x) = f(x) = 0\) with a sign change in \(f\). If \(f\) changes from positive to negative at \(x=c\) then \(F\) has a local maximum at \(x=c\). If \(f\) changes from negative to positive then \(F\) has a local minimum at that point, which mirrors the first derivative test from Unit 5.
Concavity of \(F\) is controlled by \(F''(x) = f'(x)\). Where the graph of \(f\) is increasing you have \(f'(x) > 0\) so \(F\) is concave up. Where the graph of \(f\) is decreasing you have \(f'(x) < 0\) so \(F\) is concave down, which allows you to sketch \(F\) without computing any antiderivatives.
Inflection points of \(F\) occur where \(F''(x)=f'(x)\) changes sign which is where \(f\) has a local extremum. If \(f\) peaks at \(x=p\) then \(F\) changes concavity at \(x=p\). Always confirm the sign change in \(f'(x)\) by reasoning from the local shape of \(f\) rather than only checking that \(f'(x)=0\).
The value of \(F\) at any \(x\) is the initial value \(F(a)\) plus the signed area under \(f\) from \(a\) to \(x\). If the problem gives \(F(a)=K\) you can add positive area segments and subtract negative area segments to compute \(F(x)\). This area accumulation view explains why intervals with large magnitude of \(f\) cause faster rise or fall in \(F\).
Worked example: Suppose the graph of \(f\) is above the axis on \([0,2]\) with area \(3\) and below the axis on \([2,5]\) with area \(4\) and let \(F(x)=\int_0^x f(t)\,dt\). Then \(F(2)=3\) and \(F(5)=3-4=-1\). On \([0,2]\) the function \(F\) increases and on \([2,5]\) it decreases, and if \(f\) crosses the axis at \(x=2\) then \(F\) has a local maximum there.

Area Reasoning, Sketching \(F\), and Working From Graphs or Tables

To sketch \(F\) from a graph of \(f\), mark where \(f\) is positive or negative and where it is large in magnitude. Draw \(F\) rising where \(f>0\) and falling where \(f<0\), and make the slope of \(F\) steeper where \(|f|\) is larger. Insert concavity by noting whether \(f\) is increasing or decreasing to decide if \(F\) curves upward or downward.
When the graph of \(f\) is piecewise linear or composed of simple shapes, compute exact values of \(F\) at key breakpoints using geometric areas. Triangles give \(\tfrac{1}{2}bh\), rectangles give \(bh\), and trapezoids give \(\tfrac{1}{2}(b_1+b_2)h\). Use these values as anchors for your sketch of \(F\), then connect them with shapes that match the slope and concavity rules.
From a table of \(f\) values, estimate \(F(b)-F(a)\) with numerical rules such as the trapezoidal rule \(T=\tfrac{\Delta x}{2}\left[f(x_0)+2\sum f(x_i)+f(x_n)\right]\). Use concavity cues to predict if your estimate is an overestimate or an underestimate. Justify your sign and error reasoning in a sentence so graders can award explanation credit.
Connect to the Fundamental Theorem of Calculus by remembering that \(F'(x)=f(x)\). If a question asks for where \(F\) has a horizontal tangent, find where \(f(x)=0\). If it asks where \(F\) is concave up, find where \(f\) is increasing, which is the same as \(f'(x)>0\).
Common mistakes to avoid include treating \(F(x)\) as area without sign, ignoring that regions below the axis decrease the accumulation, and forgetting that a zero of \(f\) gives a horizontal tangent for \(F\) only if \(f\) changes sign. Another frequent error is sketching \(F\) with cusps when \(f\) is continuous, but a continuous \(f\) produces a differentiable \(F\) with smooth tangents. Always check continuity of \(f\) before asserting differentiability of \(F\).
Worked example: Let \(f\) be a V shaped graph with vertex at \(x=1\) and zeros at \(x=0\) and \(x=2\), positive on \((0,2)\), and symmetric about \(x=1\). If \(F(x)=\int_0^x f(t)\,dt\), then \(F\) increases on \([0,2]\), has a horizontal tangent at \(x=0\) and \(x=2\), and a local maximum at \(x=2\) because \(f\) becomes zero and then negative for \(x>2\). The point \(x=1\) is an inflection for \(F\) because \(f\) changes from increasing to decreasing at that vertex.

Practice Problems

Problem 1 (Motion, Displacement vs. Distance, and Average Value)

A particle moves along a line with velocity \(v(t)=t^{2}-4t+3\) meters per second for \(0\le t\le 6\). The initial position is \(s(0)=10\) meters. You will compute displacement, total distance traveled, final position, and the average value of the velocity on this interval. This problem checks your ability to split integrals at zeros of \(v\) and to distinguish signed area from absolute area.
(a) Find the displacement on \([0,6]\): \(\displaystyle \int_{0}^{6} v(t)\,dt\). Explain your antiderivative choice and evaluate carefully at the bounds. Interpreting the sign of the result is part of the answer because displacement can be negative or positive depending on direction.
(b) Find the total distance traveled on \([0,6]\): \(\displaystyle \int_{0}^{6} |v(t)|\,dt\). First find the times when \(v(t)=0\) so that you can remove the absolute value by splitting the interval where the sign is known. Compute each piece as a standard definite integral and then add the magnitudes.
(c) Find the position \(s(6)\). Use the net change idea \(s(6)=s(0)+\int_{0}^{6} v(t)\,dt\). Include units and state clearly whether your value is a coordinate on the line or a distance moved.
(d) Find the average value of \(v\) on \([0,6]\): \(\displaystyle v_{\text{avg}}=\frac{1}{6-0}\int_{0}^{6}v(t)\,dt\). Then solve \(v(t)=v_{\text{avg}}\) to find times when the instantaneous velocity equals the average velocity. State all solutions in the interval and explain why the Mean Value Theorem for Integrals guarantees at least one solution.

Solution to Problem 1

(a) Antiderivative: \(\int (t^{2}-4t+3)\,dt=\tfrac{t^{3}}{3}-2t^{2}+3t+C\). The displacement is \(\left[\tfrac{t^{3}}{3}-2t^{2}+3t\right]_{0}^{6}=\left(\tfrac{216}{3}-72+18\right)-0=(72-72+18)=18\) meters. The positive value means a net movement of 18 meters to the right on the time window.
(b) Zeros of \(v\): \(t^{2}-4t+3=(t-1)(t-3)=0\) so \(t=1,3\). On \([0,1]\) and \([3,6]\) the quadratic is positive and on \([1,3]\) it is negative, which lets us remove absolute values by reversing sign on the middle segment. Compute three pieces: \(\int_{0}^{1}v\,dt=\left[\tfrac{t^{3}}{3}-2t^{2}+3t\right]_{0}^{1}=\tfrac{1}{3}-2+3=\tfrac{4}{3}\); \(\int_{1}^{3}v\,dt=\left[\dots\right]_{1}^{3}=\left(9-18+9\right)-\left(\tfrac{1}{3}-2+3\right)=0-\tfrac{4}{3}=-\tfrac{4}{3}\); \(\int_{3}^{6}v\,dt=\left[\dots\right]_{3}^{6}=18-0=18\). Total distance is \(\tfrac{4}{3}+ \left|\,-\tfrac{4}{3}\,\right|+18=\tfrac{4}{3}+\tfrac{4}{3}+18=\tfrac{8}{3}+18=\tfrac{62}{3}\) meters.
(c) Position: \(s(6)=s(0)+\int_{0}^{6}v(t)\,dt=10+18=28\) meters. This is the coordinate on the line relative to the origin used in the problem statement, not the path length traveled. The larger total distance from part (b) reflects direction changes that cancel in displacement but not in distance.
(d) Average value: \(v_{\text{avg}}=\dfrac{1}{6}\int_{0}^{6}v(t)\,dt=\dfrac{18}{6}=3\) meters per second. Solve \(t^{2}-4t+3=3\) which simplifies to \(t^{2}-4t=0\) so \(t(t-4)=0\) and the solutions in \([0,6]\) are \(t=0\) and \(t=4\). The Mean Value Theorem for Integrals guarantees at least one time where \(v(t)=v_{\text{avg}}\), and we found two such times in the interval.

Problem 2 (Accumulation, FTC, Average Value, and an Improper Integral)

Let \(f(x)=(3x+1)e^{-x}\) for \(x\ge 0\), and define the accumulation function \(F(x)=\displaystyle\int_{0}^{x}f(t)\,dt\). You will analyze \(F\) using the Fundamental Theorem of Calculus, compute an exact value of \(F(3)\), find the average value of \(f\) on \([0,3]\), and decide if \(\displaystyle\int_{0}^{\infty} f(x)\,dx\) converges. This blends symbolic integration with conceptual reasoning about accumulation.
(a) Use FTC Part 1 to find \(F'(x)\) and use derivatives to determine where \(F\) increases or is concave up. Explain the link between the signs of \(f\) and \(f'\) and the behavior of \(F\). State any interval facts you use about exponential functions and linear factors.
(b) Compute \(F(3)=\displaystyle\int_{0}^{3}(3x+1)e^{-x}\,dx\) exactly. Use integration by parts or recognize that \(\dfrac{d}{dx}\!\big((ax+b)e^{-x}\big)\) can be arranged to match \(f(x)\). Show clean bounds substitution and simplify to a single number.
(c) Find the average value of \(f\) on \([0,3]\): \(f_{\text{avg}}=\dfrac{1}{3}\displaystyle\int_{0}^{3}(3x+1)e^{-x}\,dx\). Then explain in one sentence what it means for the graph of \(y=f(x)\) on this interval. If desired, identify at least one \(c\) with \(f(c)=f_{\text{avg}}\).
(d) Determine whether \(\displaystyle\int_{0}^{\infty}(3x+1)e^{-x}\,dx\) converges and find its value if it does. Set up the limit as the upper bound tends to infinity and use the fact that \(e^{-x}\) damps any polynomial factor. Justify each limit used in your evaluation.

Solution to Problem 2

(a) By FTC Part 1, \(F'(x)=f(x)=(3x+1)e^{-x}\), which is positive for \(x\ge 0\) since \(3x+1>0\) and \(e^{-x}>0\). Therefore \(F\) increases for all \(x\ge 0\). For concavity, \(F''(x)=f'(x)=\dfrac{d}{dx}\big((3x+1)e^{-x}\big)=3e^{-x}-(3x+1)e^{-x}=(2-3x)e^{-x}\), which is positive when \(x<\tfrac{2}{3}\) and negative when \(x>\tfrac{2}{3}\). Thus \(F\) is concave up on \([0,\tfrac{2}{3})\), concave down on \((\tfrac{2}{3},\infty)\), and has an inflection at \(x=\tfrac{2}{3}\).
(b) Compute \(I=\displaystyle\int_{0}^{3}(3x+1)e^{-x}\,dx\). Use parts with \(u=3x+1\) and \(dv=e^{-x}dx\), so \(du=3dx\) and \(v=-e^{-x}\). Then \(I=[-(3x+1)e^{-x}]_{0}^{3}+\displaystyle\int_{0}^{3}3e^{-x}\,dx\). Evaluate the first bracket: at \(3\) it is \(-(10)e^{-3}\) and at \(0\) it is \(-(1)e^{0}=-1\) so the bracket contributes \(-10e^{-3}-(-1)=-10e^{-3}+1\). The second integral is \(3\left[-e^{-x}\right]_{0}^{3}=3(-e^{-3}+1)=3-3e^{-3}\). Summing gives \(I=(1-10e^{-3})+(3-3e^{-3})=4-13e^{-3}\).
(c) Average value on \([0,3]\) is \(f_{\text{avg}}=\dfrac{1}{3}I=\dfrac{1}{3}\big(4-13e^{-3}\big)=\dfrac{4}{3}-\dfrac{13}{3}e^{-3}\). This height is the level of a horizontal line whose area over \([0,3]\) matches the area under \(y=f(x)\) on the same interval. Since \(f\) is continuous, the Mean Value Theorem for Integrals guarantees at least one \(c\in[0,3]\) with \(f(c)=f_{\text{avg}}\).
(d) Evaluate the improper integral \(J=\displaystyle\int_{0}^{\infty}(3x+1)e^{-x}\,dx=\lim_{b\to\infty}\int_{0}^{b}(3x+1)e^{-x}\,dx\). From part (b) with upper bound \(b\), the antiderivative method gives \(\int_{0}^{b}(3x+1)e^{-x}\,dx=\big(4-13e^{-b}\big)\) after repeating the same steps. Taking \(b\to\infty\) makes \(e^{-b}\to 0\), so the limit is \(4\). The integral converges because the exponential decay dominates the linear factor, which is a standard test for convergence with products of polynomials and exponentials.