AP Calculus AB › Unit 6
Integration measures how much something accumulates over an interval. Unit 6 builds from Riemann sums to the definite integral and the Fundamental Theorem of Calculus, then covers antiderivatives and u-substitution. Expect to estimate integrals from tables with left, right, midpoint, or trapezoidal sums, and to analyze accumulation functions defined by integrals. Students often forget the chain rule when differentiating an integral with a variable upper limit. Other common slips are dropping the constant of integration and keeping the old bounds after a substitution.
These are taken from the Regular set, with the answer and explanation shown.
1. What is the trapezoidal sum for ∫₀² x³ dx using two subintervals of equal width?
Answer: D. Δx = 1. T = 1 · [f(0) + 2f(1) + f(2)]/2 = (0 + 2 + 8)/2 = 5.
Common trap: Choosing 4 gives the exact value ∫₀² x³ dx = 16/4. The question asks for the trapezoidal estimate. Choosing 1 or 9 gives the left or right sum.
2. If ∫₂⁵ f(x) dx = 4, what is ∫₂⁵ (3 − 2f(x)) dx?
Answer: A. Split: ∫₂⁵ 3 dx − 2∫₂⁵ f(x) dx = 3(3) − 2(4) = 9 − 8 = 1.
Common trap: Choosing −5 integrates the constant 3 as just 3, giving 3 − 8. Choosing 7 uses the upper limit 5 as the length, giving 15 − 8.
3. What is the value of ∫₀¹ (2x + 3)/(x + 1) dx?
Answer: A. Rewrite: 2x + 3 = 2(x + 1) + 1, so the integrand is 2 + 1/(x + 1). Then [2x + ln|x + 1|] from 0 to 1 = (2 + ln 2) − 0 = 2 + ln 2.
Common trap: Choosing 2 + 3ln 2 keeps the 3 from the numerator as the remainder. After removing 2(x + 1), only 1 is left over.