Difficulty: Intermediate | Prerequisites: Derivatives (see Derivatives study notes), algebra, basic trigonometry
Integration is the reverse of differentiation. An indefinite integral finds a family of antiderivatives; a definite integral computes the net area under a curve between two bounds. The main techniques at this level are the power rule run backwards, u-substitution, and Riemann sum approximations.
Integral
The mathematical operation that reverses differentiation. Represented by the elongated S symbol ∫.
In simple terms, integration finds the area under a curve, or "undoes" a derivative.
Antiderivative
A function F(x) whose derivative is f(x). If F'(x) = f(x), then F is an antiderivative of f.
Think of it as the answer to "what function, when differentiated, gives me this?"
Indefinite integral
∫ f(x) dx = F(x) + C, where C is an arbitrary constant. It represents the entire family of antiderivatives.
Definite integral
∫ from a to b of f(x) dx = F(b) − F(a). It gives a specific numerical value: the net signed area between the curve and the x-axis from x = a to x = b.
Riemann sum
An approximation of a definite integral using the areas of rectangles. The interval is divided into n subintervals, and the function is evaluated at a chosen point (left endpoint, right endpoint, or midpoint) in each subinterval.
Think of it as "stacking rectangles under the curve and adding up their areas."
U-substitution
A technique that simplifies an integral by substituting u = g(x), so that du = g'(x) dx. The integral is rewritten entirely in terms of u, integrated, then converted back to x.
In simple terms: spot a function and its derivative inside the integral, let u be that function, and the integral often becomes something standard.
Fundamental Theorem of Calculus (FTC)
Connects differentiation and integration. Part 1: d/dx [∫ from a to x of f(t) dt] = f(x). Part 2: ∫ from a to b of f(x) dx = F(b) − F(a), where F is any antiderivative of f.
Worked example (Problem 13): Estimate ∫ from 0 to 8 of [2/(2+x)] dx using n = 4 rectangles and left endpoints.
Width of each rectangle: Δx = (8 − 0)/4 = 2.
Left endpoints: x = 0, 2, 4, 6.
f(0) = 2/2 = 1, f(2) = 2/4 = 1/2, f(4) = 2/6 = 1/3, f(6) = 2/8 = 1/4.
Area ≈ 2[1 + 1/2 + 1/3 + 1/4] = 2(24 + 12 + 8 + 3)/24 = 2(47/24) = 47/12 ≈ 3.917.
Worked example (Problem 14): Evaluate ∫ (√x + 1/x + 2/x³) dx.
Rewrite: ∫ (x^(1/2) + x^(−1) + 2x^(−3)) dx.
Integrate term by term: (2/3)x^(3/2) + ln|x| + 2·x^(−2)/(−2) + C.
Result: (2/3)x^(3/2) + ln|x| − 1/x² + C.
Worked example (Problem 15): Evaluate ∫ from 3 to 0 of (2x³ − 6x²) dx.
Flip the limits and negate: = −∫ from 0 to 3 of (2x³ − 6x²) dx.
Antiderivative: x⁴/2 − 2x³.
Evaluate: −[(81/2 − 54) − 0] = −[81/2 − 54] = −[−27/2] = 27/2.
Worked example (Problem 16): Evaluate ∫ from π/6 to π/4 of cos³(t) sin(t) dt.
Let u = cos t, so du = −sin t dt, which means −du = sin t dt.
The integral becomes −∫ u³ du = −u⁴/4.
Change bounds: when t = π/6, u = cos(π/6) = √3/2; when t = π/4, u = cos(π/4) = √2/2.
Evaluate: −(1/4)[(√2/2)⁴ − (√3/2)⁴] = −(1/4)[4/16 − 9/16] = −(1/4)(−5/16) = 5/64.
Worked example (Problem 17): Evaluate ∫ from e to e² of (ln x)² / x dx.
Let u = ln x, so du = (1/x) dx.
The integral becomes ∫ u² du = u³/3.
Change bounds: when x = e, u = 1; when x = e², u = 2.
Evaluate: (2³ − 1³)/3 = 7/3.
Worked example (Problem 18): Evaluate ∫ from 0 to 4 of x√(x² + 9) dx.
Let u = x² + 9, so du = 2x dx, meaning (1/2) du = x dx.
The integral becomes (1/2)∫ u^(1/2) du = (1/2)·(2/3)u^(3/2) = u^(3/2)/3.
Change bounds: when x = 0, u = 9; when x = 4, u = 25.
Evaluate: (25^(3/2) − 9^(3/2))/3 = (125 − 27)/3 = 98/3.
Power rule for integration (n ≠ −1):
\int x^n \, dx = \frac{x^{n+1}}{n+1} + CLogarithmic integral:
\int \frac{1}{x} \, dx = \ln|x| + CFundamental Theorem of Calculus (Part 2):
\int_a^b f(x) \, dx = F(b) - F(a)U-substitution pattern: if the integrand contains g(x) and g'(x), let u = g(x).
Flipping limits reverses the sign: ∫ from a to b = −∫ from b to a.
Left-endpoint Riemann sum: Area ≈ Δx · [f(x₀) + f(x₁) + ... + f(x_{n−1})], where Δx = (b − a)/n.
Students often forget the + C on indefinite integrals. Every indefinite integral has an arbitrary constant. Definite integrals do not, because the constant cancels when you compute F(b) − F(a).
Students often confuse left-endpoint and right-endpoint sums. With left endpoints and n = 4, you evaluate f at x₀, x₁, x₂, x₃ (not x₁ through x₄).
Students often attempt u-substitution but forget to convert every part of the integrand to u. If any x remains after substitution, the substitution is incomplete or incorrect.
Students often forget to change the limits of integration when using u-substitution on a definite integral. If you switch to u, the bounds must be expressed in terms of u. Otherwise, back-substitute to x before evaluating.
⚠️ Riemann sum problems are a standard exam question. Know how to set up the width Δx and identify the correct sample points.
⚠️ U-substitution appears in nearly every integral exam problem at this level. The key skill is spotting what to let u equal: look for a function whose derivative is also present in the integrand.
⚠️ When the upper limit is smaller than the lower limit, remember to flip and negate. Forgetting this is a common source of sign errors.
⚠️ Always simplify your integrand before integrating. Rewriting √x as x^(1/2) and 1/x³ as x^(−3) makes the power rule applicable.
True or false: ∫ x³ dx = x⁴ + C.
False. The correct answer is x⁴/4 + C.
Fill in the blank: In a left-endpoint Riemann sum with n = 4 on [0, 8], the sample points are x = ______.
0, 2, 4, 6.
True or false: If you use u-substitution on a definite integral, you must change the bounds to u-values.
True (or you must back-substitute to x before evaluating).
Fill in the blank: ∫ (1/x) dx = ______.
ln|x| + C.
True or false: ∫ from 5 to 3 of f(x) dx = ∫ from 3 to 5 of f(x) dx.
False. They are negatives of each other.
Q: Estimate ∫ from 0 to 8 of [2/(2+x)] dx using n = 4 left-endpoint rectangles.
A: Δx = 2. Left endpoints: 0, 2, 4, 6. Sum = 2[1 + 1/2 + 1/3 + 1/4] = 47/12.
Q: Evaluate ∫ (√x + 1/x + 2/x³) dx.
A: (2/3)x^(3/2) + ln|x| − 1/x² + C.
Q: Evaluate ∫ from 3 to 0 of (2x³ − 6x²) dx.
A: Flip limits: = −∫ from 0 to 3. Antiderivative: x⁴/2 − 2x³. Result: 27/2.
Q: Evaluate ∫ from π/6 to π/4 of cos³(t) sin(t) dt.
A: Let u = cos t. The integral becomes −u⁴/4 evaluated from √3/2 to √2/2. Result: 5/64.
Q: Evaluate ∫ from 0 to 4 of x√(x² + 9) dx.
A: Let u = x² + 9. Result: (25^(3/2) − 9^(3/2))/3 = 98/3.
Integration is the inverse of differentiation (see the Derivatives study notes). The Fundamental Theorem of Calculus makes this relationship precise: every continuous function has an antiderivative, and the definite integral is computed by evaluating it.
Riemann sums connect to the formal definition of the integral as a limit of sums, tying this topic back to the Limits study notes.
U-substitution is the chain rule in reverse. Recognising this pattern makes both topics easier: if you are comfortable differentiating composite functions, you can spot u-substitution opportunities naturally.
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