Source: AP Calculus BC, Section I (Parts A and B)
Difficulty: Intermediate to Advanced | Prerequisites: comfort with basic antiderivatives, the power rule for integration, and the Fundamental Theorem of Calculus. You should also know how to set up a definite integral from a word problem or a graph.
Integration is the second pillar of calculus, and BC-level questions push beyond basic antiderivatives into techniques (u-substitution, partial fractions, integration by parts) and applications (arc length, volumes, average value, Riemann sums from data). Roughly a quarter to a third of the multiple-choice section tests integration in some form.
If you are coming in cold, the key prerequisite is fluency with the Fundamental Theorem of Calculus: the idea that a definite integral computes net area, and that the derivative of an accumulation function gives back the integrand. Everything below builds on that.
Master u-substitution (including changing the limits), partial fractions for rational integrands, and integration by parts. Know the arc length and average value formulas by heart. Be ready to evaluate definite integrals from graphs by computing geometric areas, and to approximate integrals using Riemann sums from a data table.
u-substitution
The integration counterpart of the chain rule. You identify an inner function u = g(x), compute du = g'(x) dx, and rewrite the integral in terms of u. For definite integrals, convert the limits to u-values too. Think of it as: undoing the chain rule.
Partial fraction decomposition
A method for integrating rational functions by splitting the fraction into simpler pieces. For example, (5x + 8) / (x^2 + 3x + 2) becomes A/(x + 1) + B/(x + 2), and each piece integrates to a natural logarithm. In simple terms, break a complicated fraction into easy ones.
Integration by parts
Based on the product rule in reverse. The formula is: integral of u dv = uv minus integral of v du. You choose u and dv from the integrand so that the new integral (v du) is simpler than the original. Think of it as: swapping one hard integral for an easier one.
Improper integral
An integral with an infinite limit of integration, or an integrand that is undefined at a point in the interval. Evaluated as a limit: replace the infinity (or the trouble point) with a variable, integrate, then take the limit. The integral either converges to a finite value or diverges.
Riemann sum
An approximation of a definite integral using rectangles. A right Riemann sum uses the function value at the right endpoint of each subinterval as the height. When given a data table, you read off the function values and multiply by the subinterval widths.
Arc length
The length of a curve y = f(x) from x = a to x = b, computed by integrating sqrt(1 + [f'(x)]^2) dx. In simple terms, you are adding up the lengths of tiny hypotenuses along the curve.
Average value of a function
The average value of f on [a, b] is (1/(b - a)) times the integral from a to b of f(x) dx. Think of it as: the height of a rectangle with the same base and the same area as the region under the curve.
Fundamental Theorem of Calculus (FTC)
Part 1: if F(x) = integral from a to x of f(t) dt, then F'(x) = f(x). Part 2: the integral from a to b of f(x) dx = F(b) minus F(a), where F is any antiderivative of f. In simple terms, differentiation and integration undo each other.
Volume by cross sections
When a solid has a known cross-sectional shape (squares, semicircles, equilateral triangles) perpendicular to an axis, the volume is the integral of the cross-sectional area along that axis.
Identify the inner function u and compute du. Rewrite every part of the integrand in terms of u and du.
For definite integrals, convert the x-limits to u-limits so you never need to back-substitute.
Example: integral from -1 to 4 of x(x^2 - 3)^5 dx with u = x^2 - 3. Then du = 2x dx, so x dx = du/2. When x = -1, u = -2. When x = 4, u = 13. The integral becomes (1/2) times integral from -2 to 13 of u^5 du.
Watch for the factor adjustment: if du = 2x dx but the integrand has x dx, you need the 1/2 factor.
Factor the denominator. Write the fraction as a sum of simpler fractions with unknown coefficients.
Solve for the coefficients by plugging in convenient x-values or by equating coefficients.
Example: (5x + 8) / [(x + 1)(x + 2)] = A/(x + 1) + B/(x + 2). Multiply both sides by (x + 1)(x + 2) to get 5x + 8 = A(x + 2) + B(x + 1). Set x = -1: 3 = A. Set x = -2: -2 = -B, so B = 2.
Each term integrates to a natural log: integral of A/(x + 1) dx = A ln|x + 1|.
For the definite integral from 0 to 1: evaluate 3 ln|x + 1| + 2 ln|x + 2| at x = 1 and x = 0, giving 3 ln 2 + 2 ln 3 - 2 ln 2 = ln 2 + 2 ln 3 = ln(18).
Use the formula: integral of u dv = uv minus integral of v du.
Choose u and dv using LIATE (Logarithmic, Inverse trig, Algebraic, Trig, Exponential) as a guide for what to call u.
Example: if integral of f(x) sin(x) dx = -f(x) cos(x) + integral of 4x^3 cos(x) dx, this matches the by-parts formula with u = f(x), dv = sin(x) dx, v = -cos(x), and du = f'(x) dx. Comparing, f'(x) = 4x^3, so f(x) = x^4.
Replace the infinite limit with a variable, integrate, then take the limit.
Example: integral from 1 to infinity of x e^(-x^2) dx. Let u = -x^2, du = -2x dx. The integral becomes -(1/2) integral of e^u du = -(1/2) e^(-x^2). Evaluate from 1 to infinity: 0 - (-(1/2)e^(-1)) = 1/(2e). The integral converges.
If the limit is finite, the integral converges. If the limit is infinite or does not exist, it diverges.
When given a table of t and R(t) values at unequally spaced points, a right Riemann sum uses the right-endpoint value times the subinterval width.
Example: given R(4) = 6.5, R(7) = 6.2, R(12) = 5.9, R(15) = 5.6, with 50 litres already in the tank at t = 4. Using three subintervals [4,7], [7,12], [12,15] with right endpoints: 6.2(3) + 5.9(5) + 5.6(3) = 18.6 + 29.5 + 16.8 = 64.9. Add the initial 50 litres: 50 + 64.9 = 114.9.
When f is defined by a graph with line segments or known shapes, compute the integral by adding signed areas.
Triangles: (1/2)(base)(height). Areas below the x-axis are negative.
Example: a graph with a line from (0, 1) to (2, 5) gives a trapezoid, and a horizontal segment from (2, -3) to (4, -3) gives a rectangle with negative area. Sum these to get the total integral.
For y = f(x) from x = a to x = b: L = integral from a to b of sqrt(1 + [f'(x)]^2) dx.
Example: y = ln(x) from x = 1 to x = 2. Since f'(x) = 1/x, [f'(x)]^2 = 1/x^2. The arc length is integral from 1 to 2 of sqrt(1 + 1/x^2) dx.
Average value of f on [a, b] = (1/(b - a)) times integral from a to b of f(x) dx.
Example: average value of sqrt(cos x) on [0, pi/2] = (2/pi) times integral from 0 to pi/2 of sqrt(cos x) dx. This requires a calculator; the result is approximately 0.763.
If the cross sections perpendicular to the x-axis are squares with side length s(x), the volume is integral of [s(x)]^2 dx.
Example: region bounded above by y = ln(3 - x) and below by y = 0, for 0 <= x <= 2. Cross sections are squares, so V = integral from 0 to 2 of [ln(3 - x)]^2 dx, which evaluates to approximately 1.029.
The integral of t^n dt = t^(n+1)/(n+1) + C, valid for n not equal to -1.
Example: integral from 1 to 4 of t^(-3/2) dt = [t^(-1/2) / (-1/2)] from 1 to 4 = -2 t^(-1/2) from 1 to 4 = -2(1/2) + 2(1) = -1 + 2 = 1. The answer is (E) 1.
\text{Arc length: } L = \int_a^b \sqrt{1 + [f'(x)]^2}\, dx\text{Average value: } f_{\text{avg}} = \frac{1}{b - a} \int_a^b f(x)\, dx\text{Integration by parts: } \int u\, dv = uv - \int v\, du\text{Volume (square cross sections): } V = \int_a^b [s(x)]^2\, dxRight Riemann sum: sum of f(right endpoint) times delta-x for each subinterval.
Riemann sums are how real measurement data (flow rates, speed readings, temperature logs) get turned into totals when you only have discrete samples rather than a continuous function. Engineers computing fluid volumes in tanks use exactly the approach in Q8. Arc length calculations appear in road design and cable engineering.
Students forget to change the limits when doing u-substitution on a definite integral. If you switch to u, the limits must also switch to u-values. Otherwise you get the wrong answer.
In partial fractions, students sometimes skip factoring the denominator completely. You need fully factored linear or irreducible quadratic factors before writing the decomposition.
Students confuse "the integral converges" with "the integrand goes to zero". An integrand can approach zero and still have a divergent integral (1/x is the classic example). Convergence of an improper integral means the limit of the antiderivative is finite.
When computing area from a graph, students often forget to subtract the area below the x-axis. Net signed area is what the definite integral computes.
u-substitution with limit conversion is tested every year. Practise until the limit-switching step is automatic.
Partial fractions appear when the integrand is a rational function whose denominator factors into distinct linear terms. The exam tends to give you the factored form or one that factors easily.
Integration by parts is a BC-only technique (not on AB). Expect one question that either asks you to evaluate a by-parts integral or to identify f(x) from a given by-parts equation.
Improper integrals: know the u-substitution approach for integrands like x e^(-x^2). The exam tests whether you can evaluate the limit and determine convergence.
Riemann sums from tables: the exam loves giving unequally spaced data and asking for a right or left sum. Read the table carefully and multiply each function value by the correct width.
Fill in the blank: if u = x^2 - 3 and du = 2x dx, then x dx = ____.
True or false: when you do u-substitution on a definite integral, you must convert the limits to u-values.
Fill in the blank: the arc length of y = f(x) from a to b is the integral of sqrt(1 + ____) dx.
True or false: integral from 1 to infinity of (1/x) dx converges.
Fill in the blank: the average value formula divides the definite integral by ____.
Answers: 1. du/2. 2. True. 3. [f'(x)]^2. 4. False (it diverges; ln(x) goes to infinity). 5. (b - a).
Q: Which integral gives the arc length of y = ln(x) from x = 1 to x = 2?
A: Integral from 1 to 2 of sqrt(1 + 1/x^2) dx. Since f'(x) = 1/x, you square it and add 1 under the square root. Answer: (A).
Q: Using u = x^2 - 3, the integral from -1 to 4 of x(x^2 - 3)^5 dx equals which of the following?
A: (1/2) integral from -2 to 13 of u^5 du. Since du = 2x dx, x dx = du/2. Limits convert: x = -1 gives u = -2, x = 4 gives u = 13. Answer: (C).
Q: A tank contains 50 litres at t = 4. Oil flows in at rate R(t). Using a right Riemann sum with three subintervals on the data R(4) = 6.5, R(7) = 6.2, R(12) = 5.9, R(15) = 5.6, approximate the amount at t = 15.
A: 114.9 litres. Right sum: 6.2(3) + 5.9(5) + 5.6(3) = 64.9. Add the initial 50. Answer: (C).
Q: Integral from 0 to 1 of (5x + 8)/(x^2 + 3x + 2) dx equals?
A: ln(18). Decompose as 3/(x + 1) + 2/(x + 2). Integrate and evaluate: 3 ln 2 + 2 ln 3 - 0 - 2 ln 2 = ln 2 + 2 ln 3 = ln(2 times 9) = ln(18). Answer: (C).
Q: Integral from 1 to infinity of x e^(-x^2) dx is?
A: 1/(2e). Substitute u = -x^2: the antiderivative is -(1/2)e^(-x^2). Evaluating from 1 to infinity gives 0 + 1/(2e). Answer: (B).
Q: What is the integral from 1 to 4 of t^(-3/2) dt?
A: 1. The antiderivative is -2 t^(-1/2). Evaluating: -2(1/2) + 2(1) = 1. Answer: (E).
Q: What is the volume of the solid with base under y = ln(3 - x) for 0 <= x <= 2, with square cross sections perpendicular to the x-axis?
A: Approximately 1.029. V = integral from 0 to 2 of [ln(3 - x)]^2 dx. Answer: (B).
u-substitution is the reverse of the chain rule studied in the derivatives notes. Integration by parts connects to the product rule. Improper integrals link to series convergence: the integral test uses improper integrals to determine whether a series converges. Riemann sums are the conceptual foundation of the definite integral itself.
Tags: u-substitution, partial fractions, integration by parts, improper integral, convergent, divergent, Riemann sum, right Riemann sum, arc length, average value, FTC, Fundamental Theorem of Calculus, volumes with cross sections, square cross sections, definite integral, area under curve, AP Calculus BC, 2012 practice exam, multiple choice, integration techniques