Difficulty: Introductory-Intermediate | Prerequisites: Calc I (derivatives, limits, basic trig identities)
Integration is the second major pillar of calculus, sitting opposite differentiation. Where derivatives ask "how fast is this changing?", integrals ask "how much has accumulated?" This topic covers the foundational definitions (definite integrals, integral functions, antiderivatives) and the core techniques you will use in every subsequent Calc II unit, from integration by parts through to series. If you are comfortable with basic derivatives, trig identities, and the power rule, you are ready for this material.
Integration reverses differentiation. A definite integral gives you a number (the accumulated quantity between two bounds), an integral function gives you a new function (one bound is a variable), and a general antiderivative gives you a family of functions differing by a constant C. Most of the work is recognising which rule or identity transforms the integrand into something you can antidifferentiate directly.
Definite integral
∫ₐᵇ f(x) dx, where a and b are both numbers. The output is a single number representing the net signed area under f(x) between x = a and x = b.
In simple terms, this means: "add up all the tiny slices of area from a to b and hand back one number."
Integral function
∫ₐʷ f(t) dt, where a is a fixed number and w is a variable. The output is a function of w, not a number. As you slide w along the x-axis, the accumulated area changes.
Think of it as: a running total that grows (or shrinks) as you move the upper limit.
General antiderivative
∫ f(x) dx = F(x) + C, where F′(x) = f(x). The "+C" accounts for the fact that infinitely many functions share the same derivative (they differ only by a vertical shift).
Think of it as: "undoing" the derivative, with an unknown constant tacked on because differentiation erases constants.
Constant of integration (C)
The arbitrary constant added to every indefinite integral. It is determined only when you have an initial condition (e.g. f′(0) = 0).
The power rule in reverse: ∫ xⁿ dx = xⁿ⁺¹ / (n+1) + C, provided n ≠ −1.
When n = −1: ∫ x⁻¹ dx = ln|x| + C.
Trig antiderivatives follow directly from memorised derivatives: ∫ cos(x) dx = sin(x) + C, ∫ sin(x) dx = −cos(x) + C, and so on.
Constants pull out freely: ∫ k·f(x) dx = k · ∫ f(x) dx.
Sums and differences split: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx.
When you are given f″(x) and a condition like f′(0) = 0, integrate once to get f′(x) + C₁, plug in the condition to pin down C₁, then integrate again to get f(x) + C₂. Each integration introduces one new constant, and each condition eliminates one.
Worked example: f″(x) = 7x + cos(x) − 2, f′(0) = 0.
First integral: f′(x) = 7x²/2 + sin(x) − 2x + C₁. Applying f′(0) = 0 gives C₁ = 0.
Second integral: f(x) = 7x³/6 − cos(x) − x² + C₂.
The power rule works for any real exponent, including irrational or transcendental ones. For ∫ x^(−sin(3)) dx, treat −sin(3) as a plain constant k and apply the rule: xᵏ⁺¹ / (k+1) + C.
Worked example: ∫ [2 − (1/8)x^(−sin(3))] dx = 2x − x^(−sin(3)+1) / [−8·sin(3) + 8] + C. (The fraction simplifies once you combine the coefficient 1/8 with the new exponent's denominator.)
Many integrands involving products of trig functions need to be rewritten using identities before you can integrate. The key identities to keep close:
tan(x) = sin(x)/cos(x)
csc²(x) = 1/sin²(x)
cos²(x) − 1 = −sin²(x)
sec(x) = 1/cos(x)
Worked example: ∫ tan(x)·csc²(x)·(cos²(x) − 1) dx.
Rewrite step by step: tan(x) = sin(x)/cos(x), csc²(x) = 1/sin²(x), cos²(x) − 1 = −sin²(x). Multiply through: [sin(x)/cos(x)] · [1/sin²(x)] · [−sin²(x)] = −sin(x)/cos(x) · 1 ... but notice that the sin²(x) terms cancel, leaving just sin(x). So the integral reduces to ∫ sin(x) dx = −cos(x) + C.
When the integrand is a fraction like (1+x)/x, split it into separate terms before integrating: (1+x)/x = 1/x + 1.
Worked example: ∫₁² (1+x)/x dx = ∫₁² (1/x + 1) dx = [ln|x| + x]₁² = (ln 2 + 2) − (ln 1 + 1) = ln 2 + 1.
When integrating |f(x)|, you must identify where f(x) changes sign on the interval, then split the integral at those roots. On sub-intervals where f(x) ≥ 0, the absolute value does nothing. Where f(x) < 0, replace |f(x)| with −f(x).
Worked example: ∫₀⁵ |x² − 4x + 3| dx.
Factor: x² − 4x + 3 = (x−1)(x−3). Roots at x = 1 and x = 3. Sign analysis on [0, 5]: positive on [0,1], negative on [1,3], positive on [3,5].
Split: ∫₀¹ (x²−4x+3) dx + ∫₁³ −(x²−4x+3) dx + ∫₃⁵ (x²−4x+3) dx. Evaluate each piece using the antiderivative x³/3 − 2x² + 3x.
Power rule (integration)
∫ xⁿ dx = xⁿ⁺¹ / (n+1) + C, for n ≠ −1
Logarithmic case
∫ 1/x dx = ln|x| + C
Basic trig antiderivatives
∫ cos(x) dx = sin(x) + C
∫ sin(x) dx = −cos(x) + C
∫ sec²(x) dx = tan(x) + C
∫ csc²(x) dx = −cot(x) + C
Trig identities used in simplification
tan(x) = sin(x) / cos(x)
csc(x) = 1 / sin(x), sec(x) = 1 / cos(x)
sin²(x) + cos²(x) = 1, so cos²(x) − 1 = −sin²(x)
Fundamental Theorem of Calculus (evaluation)
∫ₐᵇ f(x) dx = F(b) − F(a), where F is any antiderivative of f.
Students often forget the "+C" on indefinite integrals. Every indefinite integral has a family of solutions. The constant disappears only in definite integrals (it cancels in F(b) − F(a)).
Confusing the definite integral (a number) with the general antiderivative (a function + C). They are related by the Fundamental Theorem, but they are not the same object.
Applying the power rule when n = −1. The expression x⁰/0 is undefined. The antiderivative of 1/x is ln|x| + C, not x⁰/0.
Forgetting to check sign changes when integrating an absolute value. If you skip the sign analysis and just drop the absolute value bars, you will get the wrong answer on any interval where f(x) is negative.
⚠️ The power rule with non-standard exponents (fractions, negatives, irrational numbers like −sin(3)) appears frequently. If you can handle those without blinking, you are ahead of most of the class.
⚠️ Absolute value integrals are a favourite exam question because they test two skills at once: factoring to find roots and careful sign analysis on sub-intervals.
⚠️ Trig identity problems reward preparation. Memorise the fundamental identities (Pythagorean, reciprocal, quotient) so you can spot simplifications quickly under time pressure.
⚠️ Initial-condition problems (given f″, find f) test whether you can integrate twice and use conditions to pin down each constant. A common error is applying the condition to the wrong stage of the integration.
True or false: ∫₁³ f(x) dx always produces a function of x.
Fill in the blank: ∫ x⁴ dx = ______ + C.
True or false: The "+C" in an indefinite integral can be ignored when evaluating a definite integral.
Fill in the blank: ∫ sin(x) dx = ______ + C.
True or false: To integrate |x² − 4x + 3| on [0, 5], you can drop the absolute value bars and integrate directly.
Answers: 1. False (both bounds are numbers, so the result is a number). 2. x⁵/5. 3. True (C cancels in F(b) − F(a)). 4. −cos(x). 5. False (the expression changes sign on the interval, so you must split at the roots).
Q: Given f″(x) = 6x − 4 and f′(1) = 2, find f′(x).
A: Integrate f″(x): f′(x) = 3x² − 4x + C. Apply f′(1) = 2: 3(1) − 4(1) + C = 2, so C = 3. Therefore f′(x) = 3x² − 4x + 3.
Q: Evaluate ∫₀² (3/x + 2) dx. Be careful with the lower bound.
A: The integrand 3/x is undefined at x = 0, so the integral is improper. As written, it does not converge in the ordinary sense. This is a trick question testing whether you check for discontinuities before integrating.
Q: Compute ∫ sec(x)·tan(x)·cos²(x) dx.
A: Rewrite: sec(x) = 1/cos(x), so the integrand becomes [tan(x)·cos²(x)]/cos(x) = tan(x)·cos(x) = sin(x). Then ∫ sin(x) dx = −cos(x) + C.
Q: Evaluate ∫₀⁴ |x − 2| dx.
A: The expression x − 2 changes sign at x = 2. Split: ∫₀² −(x−2) dx + ∫₂⁴ (x−2) dx = ∫₀² (2−x) dx + ∫₂⁴ (x−2) dx = [2x − x²/2]₀² + [x²/2 − 2x]₂⁴ = (4−2) + (8−8 − (2−4)) = 2 + 2 = 4.
Q: Find ∫ (x³ + 1)/x² dx.
A: Split the fraction: x³/x² + 1/x² = x + x⁻². Integrate: x²/2 + x⁻¹/(−1) + C = x²/2 − 1/x + C.
This material connects directly to the Fundamental Theorem of Calculus, which you will formalise in the next unit. The FTC ties together the definite integral (area accumulation) and the antiderivative (reversing differentiation) into one theorem.
Trig identity manipulation here is the warm-up for the full integration-by-trig-substitution unit later in Calc II. The more fluent you are with identities now, the less painful that topic will be.
Absolute value integrals reappear when you study arc length and surface area of revolution, both of which involve square roots that behave like absolute values.
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