Antidifferentiation | Free AP Calculus AB Course

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<h1>AP Calculus AB: Antidifferentiation</h1>

Antidifferentiation reverses the process you've been building fluency with all along: given a derivative, find the function it came from. This free, complete lesson covers basic antiderivative formulas, u-substitution, logarithmic and arcsin/arctan forms, and applications to initial-value problems and motion along a line. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the full AP Calculus AB Question Bank linked below.

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Every antiderivative comes with an arbitrary constant of integration, C, since the derivative of any constant is 0. Two functions with the same derivative on an interval can only differ by a constant.

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<h2>What's covered in this lesson</h2>

<ol>

<li><a href="#basic-formulas">Basic Antiderivative Formulas</a></li>

<li><a href="#u-sub">U-Substitution</a></li>

<li><a href="#log-arc">Logarithmic and Arcsin/Arctan Forms</a></li>

<li><a href="#applications">Applications: Initial-Value Problems and Motion</a></li>

<li><a href="#mistakes">Common Mistakes to Avoid</a></li>

<li><a href="#faq">Frequently Asked Questions</a></li>

</ol>

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<h2 id="basic-formulas">1. Basic Antiderivative Formulas</h2>

<p>The <strong>antiderivative</strong> (or indefinite integral) of f(x) is a function F(x) whose derivative is f(x), written ∫f(x) dx = F(x) + C. A constant factor pulls out of the integral, and the integral of a sum is the sum of the integrals:</p>

<tr><th>Rule</th><th>Formula</th></tr>

<tr><td>Power Rule</td><td>∫u<sup>n</sup> du = <span class="frac"><span class="num">u<sup>n+1</sup></span><span class="den">n + 1</span></span> + C   (n ≠ −1)</td></tr>

<tr><td>Cosine</td><td>∫cos u du = sin u + C</td></tr>

<tr><td>Secant squared</td><td>∫sec² u du = tan u + C</td></tr>

<tr><td>Exponential</td><td>∫e<sup>u</sup> du = e<sup>u</sup> + C</td></tr>

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<p><strong>Worked Example:</strong> Find ∫(3x⁴ − 2x + 7) dx.</p>

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<p><span class="label">Explanation:</span> By the Power Rule: 7 · <span class="frac"><span class="num">x⁴</span><span class="den">4</span></span> + C = <span class="frac"><span class="num">7</span><span class="den">4</span></span>x⁴ + C.</p>

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<p><span class="label">Explanation:</span> Integrating term by term: 4x − 6(x²/2) + 3(x³/3) + C = 4x − 3x² + x³ + C.</p>

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<p><span class="label">Explanation:</span> The inside function 3x has derivative 3, so dividing by that factor: <span class="frac"><span class="num">1</span><span class="den">3</span></span>sin(3x) + C.</p>

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<p><span class="label">Explanation:</span> Dividing by the inside function's derivative (5): <span class="frac"><span class="num">1</span><span class="den">5</span></span>tan(5x) + C.</p>

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<p><span class="label">Explanation:</span> Dividing by the inside function's derivative (4): <span class="frac"><span class="num">1</span><span class="den">4</span></span>e<sup>4x</sup> + C.</p>

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<h2 id="u-sub">2. U-Substitution</h2>

<p>When an integral contains a function and (a multiple of) its own derivative, let u equal the "inside" function, compute du, then rewrite the whole integral in terms of u before applying a basic formula.</p>

<p><strong>Worked Example:</strong> Find ∫3(2 − 5x)⁴ dx.</p>

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<p><span class="label">Explanation:</span> u = 1 − 3x, du = −3 dx. 4∫u³(du/−3) = −<span class="frac"><span class="num">4</span><span class="den">3</span></span> · <span class="frac"><span class="num">u⁴</span><span class="den">4</span></span> + C = −<span class="frac"><span class="num">1</span><span class="den">3</span></span>(1 − 3x)⁴ + C.</p>

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<p><span class="label">Explanation:</span> u = 3x² − 1, du = 6x dx, so x dx = du/6. ∫u⁴(du/6) = <span class="frac"><span class="num">1</span><span class="den">6</span></span> · <span class="frac"><span class="num">u⁵</span><span class="den">5</span></span> + C = <span class="frac"><span class="num">1</span><span class="den">30</span></span>(3x² − 1)⁵ + C.</p>

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<p><span class="label">Explanation:</span> u = 4 − x², du = −2x dx, so x dx = du/−2. ∫u<sup>1/2</sup>(du/−2) = −<span class="frac"><span class="num">1</span><span class="den">3</span></span>u<sup>3/2</sup> + C = −<span class="frac"><span class="num">1</span><span class="den">3</span></span>(4 − x²)<sup>3/2</sup> + C.</p>

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<p><span class="label">Explanation:</span> u = sin x, du = cos x dx. ∫u² du = <span class="frac"><span class="num">u³</span><span class="den">3</span></span> + C = <span class="frac"><span class="num">1</span><span class="den">3</span></span>sin³x + C.</p>

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<p><span class="label">Explanation:</span> u = x³, du = 3x² dx, so x² dx = du/3. ∫e<sup>u</sup>(du/3) = <span class="frac"><span class="num">1</span><span class="den">3</span></span>e<sup>u</sup> + C = <span class="frac"><span class="num">1</span><span class="den">3</span></span>e<sup>x³</sup> + C.</p>

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<h2 id="log-arc">3. Logarithmic and Arcsin/Arctan Forms</h2>

<p><strong>Worked Example:</strong> Find ∫<span class="frac"><span class="num">x</span><span class="den">x² + 9</span></span> dx.</p>

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<p><span class="label">Explanation:</span> u = x² − 4, du = 2x dx, matching the numerator exactly: ∫du/u = ln|x² − 4| + C.</p>

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<p><span class="label">Explanation:</span> Here a² = 16, so a = 4: <span class="frac"><span class="num">1</span><span class="den">4</span></span>tan⁻¹<span class="frac"><span class="num">x</span><span class="den">4</span></span> + C.</p>

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<p><span class="label">Explanation:</span> Here a² = 25, so a = 5: sin⁻¹<span class="frac"><span class="num">x</span><span class="den">5</span></span> + C.</p>

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<p><span class="label">Explanation:</span> ∫tan u du = ln|sec u| + C with u = 2x; dividing by the inside derivative (2): <span class="frac"><span class="num">1</span><span class="den">2</span></span>ln|sec(2x)| + C.</p>

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<p><span class="label">Explanation:</span> u = x + 7, du = dx, so ∫du/u = ln|x + 7| + C.</p>

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<h2 id="applications">4. Applications: Initial-Value Problems and Motion</h2>

<p>Given f'(x) and a single known point (a, f(a)), antidifferentiate to find the general form (with an unknown constant C), then substitute the known point to solve for C. This same idea extends directly to motion: antidifferentiate acceleration to get velocity, and velocity to get position, using an initial condition at each stage.</p>

<p><strong>Worked Example:</strong> Find f(x) if f'(x) = 4x³ and f(1) = 6.</p>

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<p><span class="label">Explanation:</span> f(x) = 2x³ + C. f(2) = 16 + C = 10, so C = −6. f(x) = 2x³ − 6.</p>

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<p class="prompt">2. Find a curve whose slope at each point (x, y) equals <span class="frac"><span class="num">1</span><span class="den">x²</span></span>, if the curve passes through (1, 3).</p>

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<p><span class="label">Explanation:</span> dy/dx = x⁻², so y = ∫x⁻² dx = −x⁻¹ + C = −1/x + C. At (1, 3): 3 = −1 + C, so C = 4. y = −1/x + 4.</p>

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<p class="prompt">3. A particle moves along a line with velocity v(t) = 6t² − 4t. If x(0) = 5, find the position function x(t).</p>

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<p><span class="label">Explanation:</span> x(t) = ∫(6t² − 4t) dt = 2t³ − 2t² + C. x(0) = 0 + C = 5, so C = 5.</p>

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<p class="prompt">4. Using x(t) = 2t³ − 2t² + 5 from the previous problem, find the particle's position at t = 2.</p>

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<p><span class="label">Explanation:</span> x(2) = 2(8) − 2(4) + 5 = 16 − 8 + 5 = 13.</p>

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<p class="prompt">5. A particle's acceleration is a(t) = 6t − 4, and v(0) = 3. Find the velocity function v(t).</p>

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<p><span class="label">Explanation:</span> v(t) = ∫(6t − 4) dt = 3t² − 4t + C. v(0) = 0 + C = 3, so C = 3.</p>

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<h2 id="mistakes">Common Mistakes to Avoid</h2>

<li><strong>Forgetting the constant of integration, C.</strong> Every indefinite integral needs it, dropping it is one of the most common and easily avoidable point losses.</li>

<li><strong>Forgetting to divide by the "extra" constant factor when the inside function's derivative isn't exactly 1.</strong> If u = 3x, du = 3 dx, not just dx, always account for that factor before applying the basic formula.</li>

<li><strong>Choosing u incorrectly for substitution.</strong> Look for a function whose derivative (up to a constant multiple) also appears in the integral, that's the signal for what to let u equal.</li>

<li><strong>Mixing up the arcsin and arctan forms.</strong> The arcsin form has a square root and a minus sign under it (√(a² − u²)); the arctan form has no square root and a plus sign (a² + u²). Confusing these leads to entirely the wrong antiderivative.</li>

<li><strong>Substituting the initial condition before finishing the general antiderivative.</strong> Always integrate completely first, including any leftover constant multiples, before plugging in the known point to solve for C.</li>

</ul>

<h2 id="faq">Frequently Asked Questions</h2>

<h3>How can I tell when a problem needs u-substitution instead of a direct formula?</h3>

<p>If the integrand isn't a basic function of x alone, but instead involves a composite function (like (2 − 5x)⁴ or e<sup>x³</sup>) together with something resembling its derivative, that's the signal to try substitution.</p>

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<h3>Why does the antiderivative of 1/u involve an absolute value, but the derivative of ln u doesn't need one?</h3>

<p>ln u is only defined for positive u, but 1/u is defined for any nonzero u, including negative values. Using ln|u| extends the antiderivative to work correctly on both sides of 0.</p>

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<h3>What's the difference between finding f(x) from f'(x) and solving a motion problem?</h3>

<p>They use exactly the same technique, antidifferentiate, then use a given condition to solve for C. A motion problem is simply this same process applied twice in a row: once from acceleration to velocity, and again from velocity to position, each stage getting its own initial condition.</p>

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<h3>Where can I practice more problems like these?</h3>

<p>The <a href="https://theschoolofmathematics.com/quiz/antidifferentiation-quiz-1">Antidifferentiation quizzes</a> in the AP Calculus AB Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested throughout the course.</p>

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