CEGEP Calculus II — Antiderivatives and the Indefinite Integral

All 6 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Antiderivatives and the Constant of Integration

   Find the most general antiderivative of $$f(x)=6x^2-\frac{4}{x^3}+\sqrt[3]{x},$$ then verify your result by differentiating it. State the values of $x$ for which your answer is valid.

2. Q2Basic Integration Formulas

   Evaluate each indefinite integral, and state where the answer is valid.

   1. ${\displaystyle \int \frac{2x^3-5\sqrt{x}+7}{x^2}dx}$
   2. ${\displaystyle \int (3e^x+{\sec }^{2}x-\frac{2}{x})dx}$
3. Q3Integration by Substitution

   Evaluate, stating the substitution you use and checking each answer by differentiation.

   1. ${\displaystyle \int x^2\sqrt{x^3+5}dx}$
   2. ${\displaystyle \int \frac{6x}{{(3x^2+4)}^2}dx}$
4. Q4Substitution with Exponential and Trigonometric Integrands

   Evaluate each integral. In each case state the substitution, and give any restriction on $x$ that your answer needs.

   1. ${\displaystyle \int x{e}^{-x^2}dx}$
   2. ${\displaystyle \int \frac{\cos \left(\ln x\right)}{x}dx}$
   3. ${\displaystyle \int \frac{e^{2x}}{e^{2x}+1}dx}$
5. Q5Recovering a Function from Its Rate of Change

   A survey drone rises vertically from its launch pad. Its vertical velocity $t$ seconds after launch is $$v(t)=12-6\sqrt{t}\text{metres per second},0\le t\le 9,$$ and at the moment of launch the drone sits $2$ metres above the ground.

   1. Find the height $h(t)$ of the drone above the ground.
   2. The drone stops climbing when its vertical velocity first reaches zero. How high is it then?
6. Q6Synthesis — drawing on several topics in this set

   Rainwater is fed into a storage cistern at a rate $$\frac{dV}{dt}=\frac{60t}{\sqrt{t^2+9}}\text{litres per hour},$$ where $t$ is the number of hours since the storm began. The cistern already holds $100$ litres when the storm starts.

   1. Find $V(t)$, the volume of water in the cistern.
   2. How much water is in the cistern four hours into the storm, and how much of that arrived during the storm? Give both answers with units.

The 4 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.

The power rule, run backwards

Q1 asks for the most general antiderivative of a sum of powers, one written as a fraction and one as a cube root. Before anything else, rewrite every term as a power of x — a constant over x³ becomes a constant times x⁻³, and a cube root becomes an exponent of 1 ⁄ 3. The rule then applies term by term.

Raise the exponent by one, divide by the new exponent. It works for negative and fractional exponents exactly as it does for whole ones, which is the whole reason for rewriting first: there is nothing in the rule about roots or reciprocals, only about powers. Q1 then asks you to verify your result by differentiating it, and that instruction is not busywork — it is the one check in this entire topic that is guaranteed to work. If differentiating your answer does not give the integrand back, the answer is wrong, whatever you were hoping.

+ C is not decoration

An indefinite integral is not a function, it is a family of functions: every vertical shift of one antiderivative is another one, because sliding a graph up or down changes no slope anywhere. Dropping the constant does not merely lose a mark, it answers a different question from the one asked. Q1 says "the most general antiderivative", and the word general is the + C.

The constant is also the slot an initial condition fills. One antidifferentiation introduces one unknown constant, so one measurement pins it down; recover a function from its second derivative and you need two conditions, one to fix the slope and one to fix the height. Solve for the constant at the end, from the condition, and never assume it equals the starting value of the quantity — it equals whatever makes the antiderivative meet the condition, and the two agree only when the rest of the antiderivative happens to vanish at the given point.

There is no quotient rule and no product rule for integrals

Q2(a) hands you a quotient, and the only correct first move is to divide the denominator into each term of the numerator and integrate the powers that result. Integration is linear over sums and constant multiples — and over nothing else.

That is the complete list of what you may split. A quotient of integrals is not the integral of a quotient, and a product of antiderivatives is not the antiderivative of a product — differentiating a product returns two terms by the product rule, so nothing that integrates each factor on its own could ever invert it. When the integrand is a fraction, algebra comes first: divide through, split the numerator over the denominator, turn 1 ⁄ √x into a fractional power. Rewriting is a technique, not a failure to find one.

The formulas you are expected to know cold

Q2(b) is a term-by-term test of the standard list. Every entry is a derivative you already know, read from right to left.

Q2(b) also asks where the answer is valid, and the trigonometric term is the reason: sec²x and tan x are undefined at the odd multiples of π ⁄ 2, so the answer holds on an interval containing none of those, and none of x = 0 either. The habit costs nothing here and saves you later in the course, where a missed discontinuity turns an improper integral into a wrong answer rather than a missing remark.

Substitution: undoing the chain rule

Q3 and Q4 are one skill practised on different-looking integrands. Substitution reverses the chain rule, so it works precisely when the integrand contains a composite and a constant multiple of the inner function's derivative sitting there as a factor. Train yourself to look for that pair before writing anything down.

Q4 collects the three patterns worth recognising on sight. An exponential whose exponent is a function of x, accompanied by that function's derivative. A factor of 1 ⁄ x beside a function of ln x — there the 1 ⁄ x is the du. And the shape that produces a logarithm:

Whenever a numerator is a constant multiple of the derivative of its denominator, the answer is a logarithm of that denominator and no further work is needed. Q4(c) is that shape with an exponential in it, and it also shows when the absolute-value bars may be dropped: only when you can argue that the inside is positive for every x you care about. Dropping them out of habit is an error; dropping them with a stated reason is good style, and it makes the domain of the answer visible at a glance.

Recovering a function from its rate of change

Q5 and Q6 are the reason the topic exists. Q5 gives a drone's vertical velocity and its height at the moment of launch; Q6 gives the rate at which rainwater enters a cistern and the volume already in it. In both, the quantity you want is an antiderivative of the rate you were handed, and the single measurement you were handed picks one member out of the family.

Getting the most out of it

Differentiate every answer before moving on

Q1, Q3 and Q4 ask for the check in so many words, and it is worth doing on the ones that do not. It takes seconds and it catches the classic slips: the chain-rule factor that never got divided out after a substitution, the sign on ∫ sin x dx, the exponent raised instead of the coefficient divided, the dropped + C. An antiderivative is the only kind of answer in this course you can fully verify without the key.

Write the interval down as part of the answer

Four of the six questions ask where the result is valid, so treat the interval as part of the answer rather than as an afterthought. Ask what the integrand excludes — a zero denominator, a negative under an even root, a point where tan is undefined — and then which single interval your answer lives on. This is the habit the rest of Calculus II is built on.

Show the substitution, not just its result

Write "let u = …, so du = …" on its own line before touching the integral. Most substitution errors are visible in that one line and invisible everywhere else: an inside function chosen at the wrong level, a constant that was never accounted for, a stray x that the change of variable never removed.

Common questions