Source: AP Calculus BC Mock 3 (College Board 2008 PT)
Tags: integration, u-substitution, partial fractions, arc length, trapezoidal rule, Riemann sums, improper integrals, area, volume, cross-sections, AP Calculus BC
Difficulty: Intermediate to Advanced Prerequisites: Comfortable with basic antiderivatives, the Fundamental Theorem of Calculus, and basic differentiation (see Derivatives study notes).
Integration questions make up a large share of both MC and FRQ sections on the AP Calculus BC exam. Beyond simple antiderivatives, the exam tests u-substitution, partial fraction decomposition, arc length, numerical approximations (trapezoidal rule, Riemann sums), improper integrals, and applications to area and volume. The FRQ section frequently pairs integration with rates-of-change contexts where you set up the integral from a word problem.
Master u-substitution and partial fractions as your two main integration tools. Know the arc length and volume-by-cross-section formulas by heart. Be ready to set up and evaluate trapezoidal approximations from a table of values, and understand when Riemann sums converge or diverge.
u-substitution
A technique that reverses the chain rule: identify an inner function u = g(x), compute du = g'(x) dx, and rewrite the integral in terms of u. Think of it as swapping a messy integrand for a simpler one by changing variables.
Partial fraction decomposition
A method for integrating rational functions by breaking them into simpler fractions. For 8/(x^2 - 4), factor the denominator as (x-2)(x+2) and write A/(x-2) + B/(x+2). In simple terms, you split one hard fraction into two easy ones.
Arc length
The length of a curve y = f(x) from x = a to x = b, given by the integral of sqrt(1 + [f'(x)]^2) dx from a to b. Think of it as adding up tiny hypotenuses along the curve.
Trapezoidal rule
A numerical method that approximates a definite integral by summing trapezoids under the curve. For n subintervals of width h: T = (h/2)[f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n)]. In simple terms, connect the dots with straight lines and add up the areas of the resulting trapezoids.
Riemann sum
An approximation of a definite integral by summing f(x_k) * (width of subinterval) for chosen sample points x_k. Left, right, and midpoint sums are the most common. Think of it as adding up the areas of rectangles under the curve.
Improper integral
An integral with an infinite limit of integration or an integrand with an infinite discontinuity. Evaluate as a limit: the integral converges if the limit exists and is finite, and diverges otherwise.
Volume by cross-sections
If a solid has known cross-sectional area A(x) perpendicular to the x-axis, its volume is the integral of A(x) dx from a to b. Common cross-section shapes on the exam include squares, semicircles, and equilateral triangles.
For the integral of e^x / (1 + e^x) dx, let u = 1 + e^x, du = e^x dx. The integral becomes integral of du/u = ln|u| + C = ln(1 + e^x) + C.
Spotting the substitution: look for a function and its derivative sitting together in the integrand. The denominator often contains the "u" and the numerator contains "du."
For the integral of 8/(x^2 - 4) dx, factor: x^2 - 4 = (x-2)(x+2). Set 8/[(x-2)(x+2)] = A/(x-2) + B/(x+2). Solving: A = 2, B = -2. The integral becomes 2ln|x-2| - 2ln|x+2| + C = 2ln|(x-2)/(x+2)| + C.
Partial fractions apply whenever the denominator factors into distinct linear terms. If the degree of the numerator is not less than the degree of the denominator, do polynomial long division first.
For y = x^4 from x = 1 to x = 5: dy/dx = 4x^3, so (dy/dx)^2 = 16x^6. Arc length = integral from 1 to 5 of sqrt(1 + 16x^6) dx.
The formula is integral of sqrt(1 + [f'(x)]^2) dx. The common mistake is forgetting to square the derivative before adding 1.
Given f(0) = 4, f(2) = k, f(4) = 8, f(6) = 12 with 3 subintervals of width 2, the trapezoidal approximation is: (2/2)[4 + 2k + 2(8) + 12] = 4 + 2k + 16 + 12 = 32 + 2k.
If this equals 52, then 32 + 2k = 52, so k = 10.
The endpoints get weight 1, all interior points get weight 2, and everything is multiplied by h/2 where h is the subinterval width.
The integral from 0 to 1 of 1/x dx is improper because 1/x is unbounded at x = 0. As a limit: integral from epsilon to 1 of 1/x dx = ln(1) - ln(epsilon) = -ln(epsilon) -> infinity as epsilon -> 0+. The integral diverges.
When Riemann sums for an integral grow without bound as n increases, the integral diverges. The table in the exam shows sums growing past 5, 6, approaching infinity, confirming divergence.
If integral from 0 to 1 of f(x) dx = 2 and integral from 0 to 4 of f(x) dx = -3, then integral from 1 to 4 of f(x) dx = -3 - 2 = -5.
For integral from 1 to 4 of (3f(x) + 2) dx: pull out the constant and split: 3 * integral from 1 to 4 of f(x) dx + integral from 1 to 4 of 2 dx = 3(-5) + 2(3) = -15 + 6 = -9.
Region R in the first quadrant bounded by y = e^(-x^2/2), x = 3, and the axes. Cross-sections perpendicular to the x-axis are squares with side length e^(-x^2/2). Volume = integral from 0 to 3 of [e^(-x^2/2)]^2 dx = integral from 0 to 3 of e^(-x^2) dx.
This requires a calculator. The answer is approximately 0.886.
For the polar curve r = 2theta + cos(theta), 0 <= theta <= pi, the area between the curve and the x-axis is (1/2) * integral from 0 to pi of r^2 d(theta). This is a calculator problem; the answer is approximately 9.870.
The average rate of water flow over [0, 4] given R(t) = 2*sqrt(1 + 5t^3) is (1/4) * integral from 0 to 4 of R(t) dt. Evaluate with a calculator.
\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C\text{Arc length} = \int_a^b \sqrt{1 + \left[f'(x)\right]^2}\,dxT_n = \frac{h}{2}\bigl[f(x_0) + 2f(x_1) + \cdots + 2f(x_{n-1}) + f(x_n)\bigr], \quad h = \frac{b-a}{n}\text{Polar area} = \frac{1}{2}\int_\alpha^\beta r^2\,d\theta\text{Volume (cross-sections)} = \int_a^b A(x)\,dx\text{Average value} = \frac{1}{b-a}\int_a^b f(x)\,dxStudents often forget to square the derivative in the arc length formula, writing sqrt(1 + f'(x)) instead of sqrt(1 + [f'(x)]^2).
In the trapezoidal rule, a common error is giving equal weight to all function values. Remember the endpoints get weight 1 and interior points get weight 2.
When splitting a definite integral (integral from a to c = integral from a to b + integral from b to c), students sometimes get the signs wrong. If you reverse the limits, the sign flips.
For improper integrals, students sometimes evaluate the antiderivative at the trouble point directly instead of taking a limit. Always write it as a limit first.
⚠️ u-substitution with ln appears almost every year. If the integrand is f'(x)/f(x), the answer is ln|f(x)| + C.
⚠️ Partial fractions are a BC-only technique. Expect one question in Part A (no calculator).
⚠️ The trapezoidal rule from a table of values is a staple of the free-response section.
⚠️ Arc length and volume by cross-sections each appear roughly every other year. Know the setup, even if the integral requires a calculator.
⚠️ Improper integrals and divergence are tested conceptually: you need to recognise when an integral diverges and explain why.
True or false: The integral of e^x / (1 + e^x) dx equals ln(1 + e^x) + C. (True)
Fill in the blank: In the trapezoidal rule with n subintervals, the width of each subinterval is h = ______. ((b - a)/n)
True or false: The integral from 0 to 1 of 1/x dx converges. (False, it diverges)
Fill in the blank: Arc length of y = f(x) on [a, b] uses the integrand sqrt(1 + ______). ([f'(x)]^2)
True or false: 8/(x^2 - 4) can be decomposed as 2/(x-2) + 2/(x+2). (False, the second term should be -2/(x+2))
Q: What is the integral of e^x / (1 + e^x) dx?
A: ln(1 + e^x) + C. Let u = 1 + e^x, du = e^x dx.
Q: The integral of 8/(x^2 - 4) dx equals what?
A: 2 ln|(x-2)/(x+2)| + C. Use partial fractions: 8/[(x-2)(x+2)] = 2/(x-2) - 2/(x+2).
Q: What integral gives the arc length of y = x^4 from x = 1 to x = 5?
A: Integral from 1 to 5 of sqrt(1 + 16x^6) dx. Since dy/dx = 4x^3, [dy/dx]^2 = 16x^6.
Q: If f is continuous on [0, 6] with f(0) = 4, f(2) = k, f(4) = 8, f(6) = 12, and the trapezoidal approximation with 3 subintervals is 52, what is k?
A: 10. T = (2/2)[4 + 2k + 2(8) + 12] = 32 + 2k = 52, so k = 10.
Q: If integral from 0 to 1 of f(x) dx = 2 and integral from 0 to 4 of f(x) dx = -3, what is integral from 1 to 4 of (3f(x) + 2) dx?
A: -9. Integral from 1 to 4 of f(x) dx = -5. Then 3(-5) + 2(3) = -9.
Integration connects to differential equations (solving separable equations requires antidifferentiation), to parametric/polar curves (arc length and area in those coordinate systems use these same integration tools), and to series (Taylor series arise from repeated integration). The FRQ section often pairs integration with rates-of-change word problems from the derivatives topic.
u-substitution, partial fractions, arc length formula, trapezoidal rule, trapezoidal approximation, Riemann sum, left Riemann sum, right Riemann sum, improper integral, divergent integral, convergent integral, volume by cross-sections, polar area, average value of a function, definite integral properties, AP Calculus BC integration