Difficulty: Intermediate | Prerequisites: Section 13.1 (basic antiderivative rules), chain rule for derivatives.
Many integrals cannot be solved with the basic power, exponential, and log rules alone. U-substitution is the technique that handles these, and it is essentially the chain rule run in reverse. Once you have the antiderivative, the Fundamental Theorem of Calculus (FTC) connects it to definite integrals, letting you compute exact areas and accumulated quantities. Riemann sums approximate those same areas using rectangles, which is both a conceptual foundation for definite integrals and an exam topic in its own right.
U-substitution rewrites a complicated integral by letting u equal the "inner function," which turns the integral into something the basic rules can handle. Definite integrals use the FTC: find the antiderivative, evaluate at the upper bound, subtract the value at the lower bound. Left Riemann sums estimate a definite integral by summing f(xᵢ) · Δx across left endpoints.
U-substitution (integration by substitution)
A technique that reverses the chain rule. You choose an inner expression to call u, find du, and rewrite the integral entirely in terms of u. Think of it as giving the messy part of the integrand a simpler name so the basic rules apply.
Definite integral
An integral with upper and lower bounds, written ∫ₐᵇ f(x) dx. It produces a number (not a family of functions), representing the net signed area between f(x) and the x-axis from x = a to x = b.
Fundamental Theorem of Calculus (FTC)
If F(x) is an antiderivative of f(x), then ∫ₐᵇ f(x) dx = F(b) - F(a). In simple terms, to evaluate a definite integral, find the antiderivative and subtract its value at the bottom limit from its value at the top limit.
Left Riemann sum
An approximation of ∫ₐᵇ f(x) dx formed by dividing [a, b] into n equal subintervals and summing the areas of rectangles whose heights are f evaluated at each left endpoint. Think of it as stacking skinny rectangles under the curve, each one touching the curve at its left edge.
Partition
The division of the interval [a, b] into n subintervals, each of width Δx = (b - a) / n.
Subinterval width (Δx)
The width of each rectangle in a Riemann sum: Δx = (b - a) / n.
Identify an inner function inside the integrand and call it u
Compute du/dx and solve for dx (or for the remaining x-terms times dx)
Rewrite the entire integral in terms of u and du, with no x left
Integrate using basic rules, then substitute back for x
u is usually the expression inside a power, exponential, square root, or logarithm
Its derivative (or a constant multiple of it) should appear as a factor in the integrand
Common patterns:
∫ f(g(x)) · g'(x) dx: let u = g(x)
∫ (5x - 2)³ dx: let u = 5x - 2, du = 5 dx
∫ 4x · e^(x²-3) dx: let u = x² - 3, du = 2x dx
∫ (3 ln x)/x dx: let u = ln x, du = (1/x) dx
If du differs from what you have by a constant factor, adjust with multiplication
Example: if u = x² + 1 and du = 2x dx, but the integrand has x dx (not 2x dx), write x dx = (1/2) du
Find the antiderivative F(x)
Compute F(b) - F(a)
No + C needed for definite integrals, because it cancels: (F(b) + C) - (F(a) + C) = F(b) - F(a)
Two approaches: either change the limits to u-values when you substitute (no back-substitution needed), or keep x-limits, substitute back to x, then evaluate
Changing limits: if u = x² + 1 and the original limits are x = 0 to x = 2, the new limits are u = 1 to u = 5
∫ₐᵇ k·f(x) dx = k · ∫ₐᵇ f(x) dx
∫ₐᵇ [f(x) ± g(x)] dx = ∫ₐᵇ f(x) dx ± ∫ₐᵇ g(x) dx
∫ₐᵃ f(x) dx = 0
∫ₐᵇ f(x) dx = -∫ᵇₐ f(x) dx
Compute Δx = (b - a) / n
List the left endpoints: x₀ = a, x₁ = a + Δx, x₂ = a + 2Δx, ..., xₙ₋₁ = a + (n-1)Δx
Evaluate f at each left endpoint
Sum: L_n = Δx · [f(x₀) + f(x₁) + ... + f(xₙ₋₁)]
Example: f(x) = 3x² + 2x - 3 on [1, 3] with n = 5 gives Δx = 0.4, left endpoints at 1, 1.4, 1.8, 2.2, 2.6
\int f(g(x)) \cdot g'(x) \, dx = F(g(x)) + C\int_a^b f(x) \, dx = F(b) - F(a)L_n = \Delta x \cdot \sum_{i=0}^{n-1} f(x_i) \quad \text{where } \Delta x = \frac{b-a}{n}Riemann sums are how software estimates integrals numerically when no closed-form antiderivative exists, which is the basis of most computational modelling in engineering and finance. The FTC itself underpins every calculation of total distance from a speed function, total cost from a marginal cost curve, and accumulated quantity from a rate of change.
Forgetting to substitute back. After integrating in terms of u, you must replace u with the original expression in x (for indefinite integrals). Students often leave the answer in terms of u.
Leaving x-terms in a u-integral. If any x remains after substitution, either the choice of u is wrong or an algebraic step was missed. The integral must be entirely in u and du.
Adding + C to definite integrals. Definite integrals produce a number. The constant of integration cancels when you subtract F(a) from F(b).
Using right endpoints in a left Riemann sum. A left Riemann sum evaluates f at the left endpoint of each subinterval. The rightmost left endpoint is xₙ₋₁, not xₙ = b.
⚠️ Substitution problems are almost guaranteed on the exam. Practise recognising the inner function quickly.
⚠️ Definite integrals with e^(kx) and ln-based bounds (like ∫ from ln 3 to ln 5 of e^(2x) dx) are a favourite exam pattern. Simplify e^(2·ln 3) = e^(ln 9) = 9.
⚠️ Left Riemann sum problems test arithmetic as much as concept. Set up a table of endpoints and function values to stay organised.
⚠️ "Solve for the positive number b" problems give you a definite integral equation and ask you to find the upper limit. Integrate, apply the FTC, and solve the resulting equation for b.
True or false: When using u-substitution, you may leave some x-terms in the integral alongside du.
Fill in the blank: ∫₀¹ (6x⁵ + 15x⁴ - 9x² + 1) dx = F(1) - F(___), where F is the antiderivative.
True or false: A definite integral always requires + C in the final answer.
Fill in the blank: For a left Riemann sum on [2, 3] with n = 4, Δx = ___.
True or false: ∫ 16e^(-3x) dx requires u-substitution with u = -3x.
Answers: 1. False (the integral must be entirely in u). 2. F(0). 3. False (+ C cancels). 4. 0.25. 5. True.
Q: Find ∫ 16e^(-3x) dx.
A: -(16/3)e^(-3x) + C. Let u = -3x, du = -3 dx, so dx = du/(-3). The integral becomes 16 · (1/(-3)) ∫ eᵘ du = -(16/3)eᵘ + C.
Q: Find ∫ (5x - 2)³ dx.
A: (5x - 2)⁴ / 20 + C. Let u = 5x - 2, du = 5 dx. The integral becomes (1/5) ∫ u³ du = (1/5)(u⁴/4) + C.
Q: Find ∫ x(x² + 1)¹⁰ dx.
A: (x² + 1)¹¹ / 22 + C. Let u = x² + 1, du = 2x dx, so x dx = du/2. The integral becomes (1/2) ∫ u¹⁰ du.
Q: Evaluate ∫₀⁶ e^(-x+6) dx.
A: e⁶ - 1. Rewrite as e⁶ · e^(-x). The antiderivative is -e⁶ · e^(-x) = -e^(6-x). At x = 6: -e⁰ = -1. At x = 0: -e⁶. Result: -1 - (-e⁶) = e⁶ - 1.
Q: Evaluate ∫ from e³ to e⁵ of (2/x) dx.
A: 4. The antiderivative is 2 ln|x|. At x = e⁵: 2·5 = 10. At x = e³: 2·3 = 6. Result: 10 - 6 = 4.
Q: Compute the left Riemann sum for f(x) = 3x² + 2x - 3 on [1, 3] with n = 5.
A: Δx = 0.4. Left endpoints: 1, 1.4, 1.8, 2.2, 2.6. Evaluate f at each: f(1) = 2, f(1.4) = 5.68, f(1.8) = 10.32, f(2.2) = 15.92, f(2.6) = 22.48. Sum = 0.4 · (2 + 5.68 + 10.32 + 15.92 + 22.48) = 0.4 · 56.4 = 22.56.
Q: Given ∫₀ᵇ (2x² - 4) dx = 9, find the positive value of b.
A: b = 3. The antiderivative is (2/3)x³ - 4x. At x = b: (2/3)b³ - 4b. At x = 0: 0. So (2/3)b³ - 4b = 9. Multiply by 3: 2b³ - 12b = 27. Solving gives b = 3.
Q: Find ∫ 1/(2x - 5) dx.
A: (1/2) ln|2x - 5| + C. Let u = 2x - 5, du = 2 dx, so dx = du/2. The integral becomes (1/2) ∫ (1/u) du = (1/2) ln|u| + C.
Q: Find ∫ 4xe^(x² - 3) dx.
A: 2e^(x² - 3) + C. Let u = x² - 3, du = 2x dx, so 4x dx = 2 du. The integral becomes 2 ∫ eᵘ du = 2eᵘ + C.
Q: Find ∫ 15x√(-x² + 7) dx.
A: -5(-x² + 7)^(3/2) + C. Let u = -x² + 7, du = -2x dx, so 15x dx = -15/2 du. The integral becomes (-15/2) ∫ u^(1/2) du = (-15/2)(2/3)u^(3/2) + C = -5u^(3/2) + C.
Q: Find ∫ (3x² + 1)(x³ + x - 2)⁹ dx.
A: (x³ + x - 2)¹⁰ / 10 + C. Let u = x³ + x - 2, du = (3x² + 1) dx. The integrand is exactly u⁹ du. Integrate: u¹⁰/10 + C.
Q: Find ∫ (3 ln x)/x dx.
A: (3/2)(ln x)² + C. Let u = ln x, du = (1/x) dx. The integral becomes 3 ∫ u du = 3u²/2 + C.
Q: Evaluate ∫₂⁷ (x + 5/x) dx.
A: 45/2 + 5 ln(7/2). Antiderivative: x²/2 + 5 ln|x|. At x = 7: 49/2 + 5 ln 7. At x = 2: 4/2 + 5 ln 2 = 2 + 5 ln 2. Result: 49/2 - 2 + 5(ln 7 - ln 2) = 45/2 + 5 ln(7/2).
Q: Evaluate ∫₋₁¹ 5e^(3x) dx.
A: (5/3)(e³ - e⁻³). Antiderivative: (5/3)e^(3x). At x = 1: (5/3)e³. At x = -1: (5/3)e⁻³. Result: (5/3)(e³ - e⁻³).
Q: Evaluate ∫ from ln 3 to ln 5 of e^(2x) dx.
A: 8. Antiderivative: (1/2)e^(2x). At x = ln 5: (1/2)e^(2 ln 5) = (1/2)(25) = 25/2. At x = ln 3: (1/2)e^(2 ln 3) = (1/2)(9) = 9/2. Result: 25/2 - 9/2 = 8.
Q: Use a left Riemann sum with n = 4 to estimate ∫₂³ 1/(1 + 2x) dx.
A: Δx = (3 - 2)/4 = 0.25. Left endpoints: 2, 2.25, 2.5, 2.75. Evaluate: f(2) = 1/5 = 0.2, f(2.25) = 1/5.5 ≈ 0.1818, f(2.5) = 1/6 ≈ 0.1667, f(2.75) = 1/6.5 ≈ 0.1538. Sum = 0.25(0.2 + 0.1818 + 0.1667 + 0.1538) ≈ 0.25(0.7023) ≈ 0.1756.
Substitution is the bridge between the basic antiderivative rules (Section 13.1) and every applied integral in Chapter 14. Without it, most real-world integrals are unsolvable with the tools covered in this course. The FTC ties together the two halves of calculus, differentiation and integration, and Riemann sums connect to numerical methods used throughout statistics and data science.
U-substitution, integration by substitution, change of variables, definite integral, Fundamental Theorem of Calculus, FTC, left Riemann sum, Riemann approximation, partition, subinterval, Δx, net signed area, antiderivative evaluation, e^(kx) integrals, ln bounds, MAT 210, Brief Calculus, ASU, Sections 13.2, 13.3, 13.4