AP Calculus AB Unit 6 — Integration and Accumulation Worksheet

11 of the 12 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one. 11 of the 12 questions are printed below. The other 1 is built on a diagram or a table of values that does not translate to the page, so it is in the free PDF — marked below where it would have come.

1. Q1Exploring Accumulations of Change

   Meltwater runs off a glacier into a catch basin. For $0\le t\le 10$ hours it enters at a rate $r(t)$, measured in liters per hour, and the graph of $r$ consists of two line segments: one from $(0,0)$ to $(4,60)$, and one from $(4,60)$ to $(10,0)$.

   1. Using geometry, find ${\displaystyle {\int }_0^{10}r(t)dt}$.
   2. State what this number represents about the basin, with units, and explain how the units of the answer follow from the units of $r$ and of $t$.
2. Q2Approximating Areas with Riemann Sums

   This question is built around a diagram or a table of values. Open it in the PDF.

3. Q3Riemann Sums, Summation Notation and Definite Integral Notation

   Consider the limit $${lim}_{n\to \infty }{\sum }_{i=1}^n{(3+\frac{5i}{n})}^2\left(\frac{5}{n}\right).$$

   1. Identify $\Delta x$, the right endpoint $x_i$ of the $i$th subinterval, and the values of $a$ and $b$.
   2. Write the limit as a definite integral.
4. Q4The Fundamental Theorem of Calculus and Accumulation Functions

   Let $G(x)={\displaystyle {\int }_2^x(t^3-4t+1)dt}$.

   1. Find $G^{\prime }(x)$.
   2. Find $G(2)$ and $G^{\prime }(3)$.
   3. Write an equation of the line tangent to the graph of $y=G(x)$ at $x=2$.
5. Q5Interpreting the Behavior of Accumulation Functions

   The graph of a continuous function $f$ on $[0,8]$ consists of the line segments joining, in order, the points $(0,0)$, $(2,4)$, $(4,0)$, $(6,-2)$ and $(8,0)$. Define $G(x)={\displaystyle {\int }_0^{x}f(t)dt}$ for $0\le x\le 8$.

   1. Find $G(2)$, $G(4)$ and $G(6)$.
   2. Find $G(8)$.
6. Q6Applying Properties of Definite Integrals

   Suppose $f$ and $g$ are continuous and $${\int }_1^{4}f(x)dx=10,{\int }_4^{7}f(x)dx=-3,{\int }_1^{7}g(x)dx=5.$$ Evaluate each of the following.

   1. ${\displaystyle {\int }_1^{7}f(x)dx}$
   2. ${\displaystyle {\int }_7^{1}f(x)dx}$
   3. ${\displaystyle {\int }_1^7(2f(x)-3g(x))dx}$
   4. ${\displaystyle {\int }_4^{1}f(x)dx+{\int }_1^{7}g(x)dx}$
7. Q7The Fundamental Theorem of Calculus and Definite Integrals

   Evaluate each definite integral analytically, showing the antiderivative you use.

   1. ${\displaystyle {\int }_1^4(6x^2-4x+3)dx}$
   2. ${\displaystyle {\int }_1^9\frac{1}{\sqrt{x}}dx}$
8. Q8Finding Antiderivatives and Indefinite Integrals

   Find each indefinite integral.

   1. ${\displaystyle \int (4x^3-\frac{6}{x^2}+5)dx}$
   2. ${\displaystyle \int (2e^x+\frac{3}{x})dx}$
   3. ${\displaystyle \int ({\sec }^{2}x-5\sin x)dx}$
9. Q9Integrating Using Substitution
   1. Find ${\displaystyle \int 6x^2{(x^3+1)}^{3}dx}$.
   2. Evaluate ${\displaystyle {\int }_0^{3}x\sqrt{x^2+16}dx}$, converting the limits of integration to the new variable.
10. Q10Integrating Using Long Division and Completing the Square

    Each integrand must be rewritten before any antidifferentiation rule applies. Find each indefinite integral, showing the rewriting step.

    1. ${\displaystyle \int \frac{2x+7}{x+3}dx}$
    2. ${\displaystyle \int \frac{1}{x^2+6x+13}dx}$
11. Q11Selecting Techniques for Antidifferentiation

    For each integral below, name the technique you would use — a basic antiderivative rule after algebraic rewriting, substitution, long division, or completing the square — and give the feature of the integrand that told you so. Then evaluate all four integrals.

    1. ${\displaystyle \int \frac{4x}{x^2+9}dx}$
    2. ${\displaystyle \int \frac{x^2+2}{x^2+9}dx}$
    3. ${\displaystyle \int \frac{1}{x^2-2x+10}dx}$
    4. ${\displaystyle \int \frac{x-1}{\sqrt{x}}dx}$, where $x>0$
12. Q12Synthesis — drawing on several topics in this unit

    A rainwater cistern holds $50$ cubic meters of water at time $t=0$. Water enters and leaves the cistern simultaneously, so that for $0\le t\le 16$ hours the net rate at which its volume changes is $12-4\sqrt{t}$ cubic meters per hour. The volume in the cistern at time $x$ hours is therefore $$V(x)=50+{\int }_0^x(12-4\sqrt{t})dt.$$

    1. Find $V^{\prime }(x)$, and state which theorem you used.
    2. Find the time at which the volume of water is greatest, and justify that your answer gives the absolute maximum on $[0,16]$.
    3. Find the volume of water in the cistern at that time, and the volume at $x=16$. Give three-decimal accuracy where the answer is not exact.
    4. Using correct units, explain what ${\displaystyle {\int }_9^{16}(12-4\sqrt{t})dt}$ represents in this context, and state its value.

The 10 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

A definite integral of a rate is an amount

Q1 and Q12 both hand you a rate — liters per hour, cubic meters per hour — and then ask what its integral means. The answer always has the same shape: the integral of a rate of change over an interval is the net change in the quantity over that interval, and the units prove it to you.

Q1(b) is marked on exactly that sentence, so write the sentence: not "the area under the curve", which is a picture, but "the total volume of meltwater that entered the basin during the first ten hours, in liters". When the rate is drawn as straight segments you never need an antiderivative at all — the region is triangles and trapezoids, and plain geometry finishes the job.

Riemann sums: name the property, not just the direction

When a rate is only known at a handful of times, you estimate the integral by rectangles or trapezoids. An AP answer that says "the left sum is an underestimate" and stops there is half an answer. The mark is for the property of the function that forces it, and different comparisons need different properties.

One arithmetic trap goes with these: the subintervals a table gives you are often not the same width. There is no single Δt to factor out. Each rectangle uses its own width, and each trapezoid contributes the average of its two endpoint values times its own width. Write the widths down before you write the sum.

Reading a limit of sums backwards into an integral

Q3 gives you a limit of Riemann sums and wants the definite integral it converges to. Nothing here is guesswork; the sum tells you every piece if you take it apart in order.

The Fundamental Theorem runs in two directions

Students meet two statements under one name and blur them. They do opposite jobs, and the sheet asks for both — Q7 uses the first, Q4 and Q12(a) use the second.

The first turns an antiderivative into a number. The second says that an accumulation function is an antiderivative of its integrand — differentiating undoes the integrating, and you simply replace the dummy variable by the upper limit. Nothing else changes, and no antiderivative ever has to be found.

Two small facts fall straight out of the definition and both are worth a mark. First, F(a) = ∫ from a to a of f(t) dt = 0 because the limits agree — which is why Q4 can ask for a tangent line at the lower limit without giving you a value of F. Second, the letter inside must differ from the letter in the limit: you write F(x) = ∫ from a to x of f(t) dt, with t as the dummy variable. The integral has already used up its variable — t is consumed by the integration and is gone from the answer, while x survives as the input of F. Writing ∫ from a to x of f(x) dx uses one letter for two different things and makes the next line unreadable.

Reading an accumulation function off the graph of f

In Q5 you are given a graph of f built from line segments and asked for values of F(x) = ∫ from 0 to x of f(t) dt. Each value is a running signed area: add the area of each piece as you sweep right, counting anything below the axis as negative. Do it one segment at a time and keep the running total — F(6) is F(4) plus whatever happens on [4, 6], not a fresh computation.

Q12(b) is the same reasoning wearing a context. To locate the maximum of an accumulating quantity, find where its rate crosses from positive to negative, then justify: the quantity rises while the rate is positive and falls once it is negative, so that crossing gives the absolute maximum on the interval. Backing it with a candidates test — the critical point and both endpoints, values compared — is what makes the justification complete.

The properties that let you move limits around

Q6 gives you three integrals and asks for four more, with no function ever named. That is the point: these are structural rules, and you evaluate by rearranging, not by integrating.

What linearity does not cover is the absolute value: ∫ from a to b of |f(x)| dx is not | ∫ from a to b of f(x) dx | The integral of |f| reflects every negative piece upward and measures total area, in which nothing cancels; the absolute value of the integral takes the net signed area and only then drops its sign. So an integrand wrapped in absolute-value bars is never integrated straight through: split the interval at every point where the inside changes sign, integrate each piece with the sign it actually has, and add. And use the obvious check — an integral of a non-negative integrand cannot come out negative.

Choosing an antidifferentiation technique

Q8, Q9, Q10 and Q11 build one skill: looking at an integrand and knowing which move to make before trying anything. Q11 asks you to name the technique and the feature that settled it, so practise the diagnosis out loud.

Two habits finish the section. Every indefinite integral ends in + C — it is part of the answer, not decoration, and Q8 loses marks without it. A definite integral never carries one, because the constant cancels in F(b) − F(a). And every antiderivative can be checked in one line by differentiating it: if you do not get the integrand back, you have your answer before the exam does.

Preview all 9 pages

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Getting the most out of it

Answer the "why" in the language of the property

Several questions here ask for a justification, and the justification is where the marks sit. Train the two sentences: "the left sum underestimates because the rate is increasing, so every left endpoint gives the smallest value on its subinterval", and "the trapezoidal sum overestimates because the rate is concave up, so the graph lies below each chord". Direction plus property, every time.

Differentiate your antiderivative before moving on

Every indefinite integral on this sheet can be verified in one line. It costs ten seconds and it catches the classic slips: the missing chain-rule factor, the sign on ∫ sin x dx, the missing 1⁄a on an arctan, the dropped + C.

Interpret every number in a sentence with units

On the accumulation questions, once you have the value, translate it: "the cistern held about that many cubic meters at hour nine". A stated interpretation with correct units is a mark of its own on the AP exam, and it is also the fastest check you have — an interpretation that sounds absurd usually is.

Common questions