AP Calculus AB — Practice Set 1

Limits, derivatives, applications of differentiation, integration and the Fundamental Theorem — including a free-response question scored the way AP scores them.

How to use this: on free response, AP awards points for setup and for notation separately from the answer. Writing $\int_0^3 f(x)\,dx$ correctly earns a point even if the arithmetic then goes wrong. Show the integral or derivative you intend to evaluate before evaluating it.

Limits

1. Evaluate $\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$

Direct substitution gives 0/0 — indeterminate, so factor.

 x² - 4      (x - 2)(x + 2)
--------- = ----------------- = x + 2      (for x ≠ 2)
 x - 2           x - 2

lim (x + 2) = 4
x→2

Answer: 4

The routine: substitute first. If you get $0/0$, factor, rationalise, or use a known limit. Reaching for L'Hôpital's rule immediately is slower and, on the no-calculator section, more error-prone.

2. Evaluate $\displaystyle\lim_{x \to \infty} \frac{3x^2 + 5x}{2x^2 - 1}$

Divide numerator and denominator by the highest power, x²:

 3 + 5/x        3 + 0        3
------------ → --------- = -----
 2 - 1/x²        2 - 0        2

Answer: $\frac{3}{2}$

The shortcut: when the degrees are equal, the limit is the ratio of leading coefficients. If the numerator's degree is lower, the limit is 0; if higher, the limit diverges.

Derivatives

3. Find $\dfrac{dy}{dx}$ for $y = x^3 \sin x$

Product rule:  (uv)' = u'v + uv'

u = x³        u' = 3x²
v = sin x     v' = cos x

dy/dx = 3x² sin x + x³ cos x

Answer: $3x^2\sin x + x^3\cos x$

4. Find $\dfrac{dy}{dx}$ for $y = e^{2x^2}$

Chain rule:  d/dx [e^u] = e^u · u'

u  = 2x²
u' = 4x

dy/dx = 4x e^(2x²)

Answer: $4xe^{2x^2}$

Where this is lost: omitting the inner derivative and writing $e^{2x^2}$. The chain rule is the most-tested and most-forgotten rule on the exam.

5. Find $\dfrac{dy}{dx}$ if $x^2 + y^2 = 25$

Differentiate both sides with respect to x, treating y as a function of x:

2x + 2y (dy/dx) = 0

dy/dx = -x/y

Answer: $-\dfrac{x}{y}$

Implicit differentiation is signalled by an equation you cannot easily solve for $y$. It is normal and correct for the answer to contain both $x$ and $y$.

6. A particle moves with position $s(t) = t^3 - 6t^2 + 9t$. Find when it is at rest.

At rest means velocity = 0.

v(t) = s'(t) = 3t² - 12t + 9
             = 3(t² - 4t + 3)
             = 3(t - 1)(t - 3)

v(t) = 0  ⟹  t = 1  or  t = 3

Answer: $t = 1$ and $t = 3$

The vocabulary AP uses, and expects you to translate:

Speeding up is not the same as accelerating, and the distinction is examined regularly.

Applications

7. Find the absolute maximum of $f(x) = x^3 - 3x$ on $[-2, 2]$.

f'(x) = 3x² - 3 = 3(x² - 1) = 0  ⟹  x = ±1

Evaluate at critical points AND endpoints:

f(-2) = -8 + 6  = -2
f(-1) = -1 + 3  =  2      ← maximum
f(1)  =  1 - 3  = -2
f(2)  =  8 - 6  =  2      ← maximum (tied)

Absolute maximum = 2, attained at x = -1 and x = 2.

Answer: 2

The Candidates Test: on a closed interval you must check the endpoints as well as the critical points. Forgetting the endpoints is the standard error, and here it would have missed one of the two locations.

Integration

8. Evaluate $\displaystyle\int_0^2 (3x^2 + 2x)\,dx$

Antiderivative:  x³ + x²

Evaluate:  [x³ + x²]₀²
        =  (8 + 4) - (0 + 0)
        =  12

Answer: 12

9. Evaluate $\displaystyle\int 2x\sqrt{x^2+1}\,dx$

Substitute u = x² + 1,  du = 2x dx

∫ √u du  =  (2/3) u^(3/2) + C

Back-substitute:

= (2/3)(x² + 1)^(3/2) + C

Answer: $\frac{2}{3}(x^2+1)^{3/2} + C$

Recognising u-substitution: look for a function and a constant multiple of its derivative in the same integrand. Here $x^2+1$ appears with $2x$, which is exactly its derivative.

Do not omit $+C$ on an indefinite integral. It is a scoring point on free response.

Free response

10. Let $f$ be continuous with $\displaystyle\int_0^4 f(x)\,dx = 10$ and $\displaystyle\int_0^2 f(x)\,dx = 3$.

(a) Find $\int_2^4 f(x)\,dx$. (b) Find $\int_0^4 3f(x)\,dx$. (c) If $g(x) = \int_0^x f(t)\,dt$, find $g'(2)$ given $f(2) = 5$. (6 points)

(a) — 2 points

By additivity over adjacent intervals:

∫₀⁴ = ∫₀² + ∫₂⁴

10 = 3 + ∫₂⁴

∫₂⁴ f(x) dx = 7

1 point for stating the additivity property, 1 for the answer.

(b) — 2 points

Constants factor out of an integral:

∫₀⁴ 3f(x) dx = 3 ∫₀⁴ f(x) dx = 3(10) = 30

1 point for factoring the constant out, 1 for the answer.

(c) — 2 points

By the Fundamental Theorem of Calculus, Part 1:

     d    ⌠ˣ
g'(x) = ---- │ f(t) dt  =  f(x)
        dx   ⌡₀

∴ g'(2) = f(2) = 5

1 point for citing the Fundamental Theorem, 1 for the value.

Why (c) is the discriminating part: many candidates try to evaluate the integral. The Fundamental Theorem says the derivative of an accumulation function is simply the integrand evaluated at the upper limit — no integration is needed at all.

What AP Calculus rewards

1. Setup earns points independently. Write the integral or derivative you intend to compute before computing it. If the arithmetic fails, the setup point survives.
2. Notation is scored. $dx$, limits of integration, $+C$, and correct function notation each carry weight.
3. Justify with a named theorem. "By the Fundamental Theorem…", "By the Mean Value Theorem…" — justification questions require the name.
4. Units and interpretation in context. When a question is set in a real situation, the answer needs units and a sentence saying what it means.
5. Know which section allows a calculator. Do not spend no-calculator time on arithmetic you can restructure away.

Where to go next

• AP Biology Practice Set 1 — if you are sitting multiple AP exams
• SAT Math Practice Set 1 — the algebra underneath these techniques
• A-Level Mathematics Practice Paper 1 — the same calculus in the UK system

Ask StudyAI to mark your free-response working against the point allocations shown here.