Applications of Integration | Free AP Calculus AB Course

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<h1>AP Calculus AB: Applications of Integration</h1>

This is where integration becomes a tool for measuring the world: areas, volumes, distances, and accumulated change. This free, complete lesson covers area under a curve and using symmetry, area between curves, volumes with known cross sections, volumes of revolution using disks and washers, motion along a line (distance vs. displacement), and net change from a rate. 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.

</p>

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

<ol>

<li><a href="#area-curve">Area Under a Curve and Using Symmetry</a></li>

<li><a href="#area-between">Area Between Curves</a></li>

<li><a href="#cross-sections">Volumes with Known Cross Sections</a></li>

<li><a href="#disks">Volumes of Revolution: Disks</a></li>

<li><a href="#washers">Volumes of Revolution: Washers</a></li>

<li><a href="#motion">Motion Along a Line: Distance vs. Displacement</a></li>

<li><a href="#net-change">Net Change: Definite Integral of a Rate</a></li>

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

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

</ol>

</nav>

<h2 id="area-curve">1. Area Under a Curve and Using Symmetry</h2>

<p>The area bounded by y = f(x), the x-axis, and the vertical lines x = a and x = b (with f(x) ≥ 0) is ∫<sub>a</sub><sup>b</sup> f(x) dx. When a region is symmetric about an axis, integrating over half the region and doubling the result is often simpler. Area can also be found with respect to y, integrating x = g(y) between horizontal bounds.</p>

<p><strong>Worked Example:</strong> Find the area under f(x) = x² + 2 from x = 0 to x = 3.</p>

</div>

<p class="prompt">1. Find the area bounded by f(x) = 4 − x² and the x-axis, using symmetry about the y-axis.</p>

</div>

<p><span class="label">Explanation:</span> The curve crosses the x-axis at x = ±2. By symmetry: 2∫<sub>0</sub><sup>2</sup>(4 − x²) dx = 2[4x − x³/3]<sub>0</sub><sup>2</sup> = 2(16/3) = 32/3.</p>

</div>

<p class="prompt">2. Find the area under f(x) = 3x² from x = 1 to x = 4.</p>

</div>

<p><span class="label">Explanation:</span> ∫<sub>1</sub><sup>4</sup>3x² dx = [x³]<sub>1</sub><sup>4</sup> = 64 − 1 = 63.</p>

</div>

<p class="prompt">3. The region bounded by y = cos(2x) and the x-axis on [−π/4, π/4] is symmetric about the y-axis. If the area to the right of the y-axis is <span class="frac"><span class="num">1</span><span class="den">2</span></span>, what is the total area?</p>

</div>

<p><span class="label">Explanation:</span> By symmetry, the total area is twice the area of one half: 2 × 1/2 = 1.</p>

</div>

<p class="prompt">4. Find the area bounded by x = y², the y-axis, and the lines y = 0 and y = 3.</p>

</div>

<p><span class="label">Explanation:</span> Integrating with respect to y: ∫<sub>0</sub><sup>3</sup>y² dy = [y³/3]<sub>0</sub><sup>3</sup> = 27/3 = 9.</p>

</div>

<h2 id="area-between">2. Area Between Curves</h2>

<p>To find the area between y = f(x) and y = g(x), first find where they intersect, then integrate (top − bottom) between those intersection points. If which curve is on top changes, split the integral at the crossing point.</p>

<p><strong>Worked Example:</strong> Find the area between y = 2x and y = x² (they intersect at x = 0 and x = 2).</p>

</div>

<p class="prompt">1. Find the area between y = x + 2 and y = x² (they intersect at x = −1 and x = 2).</p>

</div>

<p><span class="label">Explanation:</span> The line lies above the parabola on this interval. ∫<sub>−1</sub><sup>2</sup>[(x + 2) − x²] dx = [x²/2 + 2x − x³/3]<sub>−1</sub><sup>2</sup> = 10/3 − (−7/6) = 9/2.</p>

</div>

<p class="prompt">2. Find the area between y = 6 − x² and y = x (they intersect at x = −3 and x = 2).</p>

</div>

<p><span class="label">Explanation:</span> ∫<sub>−3</sub><sup>2</sup>[(6 − x²) − x] dx = [6x − x³/3 − x²/2]<sub>−3</sub><sup>2</sup> = 22/3 − (−27/2) = 125/6.</p>

</div>

<p class="prompt">3. Two curves intersect such that f(x) ≥ g(x) on [a, c] and g(x) ≥ f(x) on [c, b]. Write the total area between them from a to b.</p>

</div>

<p><span class="label">Explanation:</span> Each piece must use (top − bottom) for that piece specifically, so the integral splits at the crossing point c with the order reversed on each side.</p>

</div>

<p class="prompt">4. Why is it essential to determine which curve is on top before setting up an area-between-curves integral?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p2-4',true)">A) So the integrand (top minus bottom) gives a positive area</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p2-4',false)">B) It doesn't matter, the integral gives the same value either way</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p2-4',false)">C) Only the leftmost curve matters</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p2-4',false)">D) The steeper curve must always be listed first</button>

</div>

<p><span class="label">Explanation:</span> Reversing the order (bottom − top) produces a negative value; the integrand must be top minus bottom to correctly represent a positive area.</p>

</div>

<h2 id="cross-sections">3. Volumes with Known Cross Sections</h2>

<p>When the area of a cross section perpendicular to an axis is known in terms of x, the volume is ∫<sub>a</sub><sup>b</sup> A(x) dx, where A(x) is the cross-sectional area formula.</p>

<p><strong>Worked Example:</strong> A solid has a base bounded by the circle x² + y² = 9, with cross sections perpendicular to the x-axis that are squares. Find the volume.</p>

<p>The square's side is 2y = 2√(9 − x²), so A(x) = 4(9 − x²).</p>

</div>

<p class="prompt">1. A solid has a base bounded by the circle x² + y² = 4, with square cross sections perpendicular to the x-axis. Find the volume.</p>

</div>

<p><span class="label">Explanation:</span> Side = 2√(4 − x²), A(x) = 4(4 − x²). V = 2∫<sub>0</sub><sup>2</sup>4(4 − x²) dx = 8[4x − x³/3]<sub>0</sub><sup>2</sup> = 8(16/3) = 128/3.</p>

</div>

<p class="prompt">2. A solid's base is bounded by y = x² and y = 4, with semicircular cross sections whose diameter stretches from the parabola to the line. Find the radius as a function of x.</p>

</div>

<p><span class="label">Explanation:</span> The diameter equals the vertical distance between the curves, 4 − x², so the radius is half that.</p>

</div>

<p class="prompt">3. For a solid whose cross sections perpendicular to the x-axis are equilateral triangles with side length s(x), what is the area formula for each cross section?</p>

</div>

<p><span class="label">Explanation:</span> This is the standard area formula for an equilateral triangle with side length s.</p>

</div>

<p class="prompt">4. Set up (but do not evaluate) the integral for the volume of a solid whose base is bounded by y = √x, the x-axis, and x = 4, with square cross sections perpendicular to the x-axis.</p>

</div>

<p><span class="label">Explanation:</span> The square's side is y = √x, so A(x) = (√x)² = x.</p>

</div>

<h2 id="disks">4. Volumes of Revolution: Disks</h2>

<p>When a region bounded by a curve and an axis is revolved about that axis, each cross section is a disk. V = π∫<sub>a</sub><sup>b</sup> r² dx (or dy), where r is the distance from the axis to the curve.</p>

<p><strong>Worked Example:</strong> Find the volume when the region bounded by y = x², x = 2, y = 0 is rotated about the x-axis.</p>

</div>

<p class="prompt">1. Find the volume when the region bounded by y = √x, x = 4, y = 0 is rotated about the x-axis.</p>

</div>

<p><span class="label">Explanation:</span> V = π∫<sub>0</sub><sup>4</sup>(√x)² dx = π∫<sub>0</sub><sup>4</sup>x dx = π[x²/2]<sub>0</sub><sup>4</sup> = 8π.</p>

</div>

<p class="prompt">2. Find the volume when the region bounded by y = x², y = 0, x = 3 is rotated about the y-axis (integrate in y, using radius x = √y, for y from 0 to 9).</p>

</div>

<p><span class="label">Explanation:</span> V = π∫<sub>0</sub><sup>9</sup>y dy = π[y²/2]<sub>0</sub><sup>9</sup> = 81π/2.</p>

</div>

<p class="prompt">3. A region bounded by y = x³, x = 0, y = 8 is rotated about the y-axis. Using radius x = y<sup>1/3</sup>, set up the integral for the volume.</p>

</div>

<p><span class="label">Explanation:</span> V = π∫(y<sup>1/3</sup>)² dy = π∫y<sup>2/3</sup> dy.</p>

</div>

<p class="prompt">4. Using disks, find the volume of a sphere of radius 3 generated by rotating a semicircle of radius 3 about the x-axis.</p>

</div>

<p><span class="label">Explanation:</span> V = (4/3)πr³ = (4/3)π(27) = 36π, matching the standard sphere volume formula.</p>

</div>

<h2 id="washers">5. Volumes of Revolution: Washers</h2>

<p>When the revolved region is bounded away from the axis (leaving a hole), each cross section is a washer: V = π∫<sub>a</sub><sup>b</sup>(R² − r²) dx, where R is the outer radius and r is the inner radius.</p>

<p><strong>Worked Example:</strong> Find the volume when the region between y = x² and y = 4 is rotated about the x-axis (they intersect at x = ±2).</p>

</div>

<p class="prompt">1. Find the volume when the region between y = x and y = x² (on [0, 1]) is rotated about the x-axis.</p>

</div>

<p><span class="label">Explanation:</span> Outer radius R = x, inner radius r = x². V = π∫<sub>0</sub><sup>1</sup>(x² − x⁴) dx = π(1/3 − 1/5) = 2π/15.</p>

</div>

<p class="prompt">2. Find the volume when the region between y = x² and y = 9 is rotated about the x-axis (they intersect at x = ±3).</p>

</div>

<p><span class="label">Explanation:</span> R = 9, r = x². V = 2π∫<sub>0</sub><sup>3</sup>(81 − x⁴) dx = 2π[81x − x⁵/5]<sub>0</sub><sup>3</sup> = 2π(972/5) = 1944π/5.</p>

</div>

<p class="prompt">3. In the washer method, if R is the outer radius and r is the inner radius, what is the formula for the volume element ΔV?</p>

</div>

<p><span class="label">Explanation:</span> A washer's volume is the outer disk's volume minus the inner disk's volume: πR²Δx − πr²Δx = π(R² − r²)Δx.</p>

</div>

<p class="prompt">4. The region bounded by y = x² and y = 2x (intersecting at x = 0 and x = 2, with y = 2x above y = x²) is rotated about the x-axis. Which radius corresponds to y = 2x?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p5-4',true)">A) R (the outer radius)</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-4',false)">B) r (the inner radius)</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-4',false)">C) Neither, both are equal</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-4',false)">D) It depends on the value of x</button>

</div>

<p><span class="label">Explanation:</span> Since y = 2x lies farther from the x-axis than y = x² on this interval, it forms the outer radius R.</p>

</div>

<h2 id="motion">6. Motion Along a Line: Distance vs. Displacement</h2>

<p><strong>Displacement</strong> (net change in position) is ∫<sub>a</sub><sup>b</sup>v(t) dt. <strong>Total distance traveled</strong> is ∫<sub>a</sub><sup>b</sup>|v(t)| dt, found by splitting the interval wherever v(t) changes sign and summing the absolute value of each piece.</p>

<p><strong>Worked Example:</strong> A particle has v(t) = t² − 4t + 3 on [0, 3]. Find the displacement and total distance traveled.</p>

<p>Displacement: ∫<sub>0</sub><sup>3</sup>(t² − 4t + 3) dt = [t³/3 − 2t² + 3t]<sub>0</sub><sup>3</sup> = 0.</p>

<p>v(t) = (t − 1)(t − 3): positive for t < 1, negative for 1 < t < 3.</p>

<p class="step-math">Distance = ∫<sub>0</sub><sup>1</sup>v dt − ∫<sub>1</sub><sup>3</sup>v dt = <span class="frac"><span class="num">4</span><span class="den">3</span></span> − (−<span class="frac"><span class="num">4</span><span class="den">3</span></span>) = <span class="frac"><span class="num">8</span><span class="den">3</span></span></p>

</div>

<p class="prompt">1. A particle has v(t) = 3t² − 6t on [0, 3]. Find the displacement from t = 0 to t = 3.</p>

</div>

<p><span class="label">Explanation:</span> ∫<sub>0</sub><sup>3</sup>(3t² − 6t) dt = [t³ − 3t²]<sub>0</sub><sup>3</sup> = 27 − 27 = 0.</p>

</div>

<p class="prompt">2. Using the same velocity v(t) = 3t² − 6t = 3t(t − 2), find the total distance traveled from t = 0 to t = 3.</p>

</div>

<p><span class="label">Explanation:</span> v < 0 on (0, 2) and v > 0 for t > 2. ∫<sub>0</sub><sup>2</sup>v dt = −4, ∫<sub>2</sub><sup>3</sup>v dt = 4. Distance = |−4| + |4| = 8.</p>

</div>

<p class="prompt">3. If v(t) ≥ 0 for all t on [a, b], how does the total distance traveled compare to the displacement?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p6-3',true)">A) They are equal</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-3',false)">B) Distance is always greater</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-3',false)">C) Displacement is always greater</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-3',false)">D) They are always opposite in sign</button>

</div>

<p><span class="label">Explanation:</span> When velocity never goes negative, the particle never reverses direction, so distance and displacement coincide exactly.</p>

</div>

<p class="prompt">4. Why does total distance traveled require ∫|v(t)| dt rather than just ∫v(t) dt?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p6-4',true)">A) ∫|v(t)| dt accumulates ground covered regardless of direction, while ∫v(t) dt allows positive and negative motion to cancel</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-4',false)">B) They are actually the same formula written differently</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-4',false)">C) ∫v(t) dt only works for constant velocity</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-4',false)">D) ∫|v(t)| dt is only used when velocity is always positive</button>

</div>

<p><span class="label">Explanation:</span> Displacement can be small or zero even if the particle traveled a long path back and forth; taking the absolute value before integrating prevents that cancellation.</p>

</div>

<h2 id="net-change">7. Net Change: Definite Integral of a Rate</h2>

<p>If f(t) = F'(t), then ∫<sub>a</sub><sup>b</sup>f(t) dt = F(b) − F(a) is the <strong>net change</strong> (or total accumulation) in the quantity F over [a, b]. This principle applies to virtually any rate: population growth, spreading rumors, marginal cost, water flow, and more.</p>

<p><strong>Worked Example:</strong> A chemical decomposes at a rate of R(t) = 5e<sup>−0.5t</sup> grams per minute. Find the amount decomposed during the first 4 minutes.</p>

</div>

<p class="prompt">1. A population grows at a rate of G(t) = 50e<sup>0.1t</sup> organisms per day. Find the increase in population from t = 0 to t = 10.</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p7-1',true)">A) 500(e − 1) ≈ 859.1 organisms</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-1',false)">B) 50(e − 1) ≈ 85.9 organisms</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-1',false)">C) 500e ≈ 1359.1 organisms</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-1',false)">D) 50e¹⁰ organisms</button>

</div>

<p><span class="label">Explanation:</span> ∫<sub>0</sub><sup>10</sup>50e<sup>0.1t</sup> dt = [500e<sup>0.1t</sup>]<sub>0</sub><sup>10</sup> = 500e − 500 = 500(e − 1) ≈ 859.1.</p>

</div>

<p class="prompt">2. The marginal cost of production is C'(x) = 2 + 0.01x dollars per unit. Find the change in cost when production increases from 100 to 200 units.</p>

</div>

<p><span class="label">Explanation:</span> ∫<sub>100</sub><sup>200</sup>(2 + 0.01x) dx = [2x + 0.005x²]<sub>100</sub><sup>200</sup> = 600 − 250 = 350.</p>

</div>

<p class="prompt">3. If f(t) represents the rate (in people per week) at which a rumor spreads, what does ∫<sub>2</sub><sup>5</sup>f(t) dt represent?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p7-3',true)">A) The number of new people who hear the rumor from week 2 to week 5</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-3',false)">B) The total number of people who have ever heard the rumor</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-3',false)">C) The rate of spread at week 5</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-3',false)">D) The average rate of spread over the 5 weeks</button>

</div>

<p><span class="label">Explanation:</span> The definite integral of a rate over an interval gives the net accumulation during that specific interval, here, the new people reached between weeks 2 and 5.</p>

</div>

<p class="prompt">4. Water flows into a tank at a rate of R(t) = 20 − 2t gallons per minute for 0 ≤ t ≤ 10. Find the total amount of water that flows in during this time.</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p7-4',true)">A) 100 gallons</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-4',false)">B) 200 gallons</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-4',false)">C) 20 gallons</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-4',false)">D) 180 gallons</button>

</div>

<p><span class="label">Explanation:</span> ∫<sub>0</sub><sup>10</sup>(20 − 2t) dt = [20t − t²]<sub>0</sub><sup>10</sup> = 200 − 100 = 100.</p>

</div>

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

<li><strong>Forgetting to find intersection points before setting up an area-between-curves integral.</strong> The bounds of integration come directly from where the curves meet, always solve for these first.</li>

<li><strong>Mixing up the radius in a cross-section problem.</strong> For a square cross section, the volume element uses (side)², not the side length itself, don't forget to square the expression for the side.</li>

<li><strong>Confusing disks and washers.</strong> Use disks when the region touches the axis of revolution (no hole); use washers when there's a gap between the region and the axis (a hole in the middle).</li>

<li><strong>Forgetting the absolute value when finding total distance traveled.</strong> ∫v(t) dt gives displacement, not distance, always check for sign changes in v(t) and split the integral accordingly when distance is asked for.</li>

<li><strong>Not connecting net change problems back to the Fundamental Theorem of Calculus.</strong> Any "rate of ___" problem asking for total accumulated change over an interval is simply ∫<sub>a</sub><sup>b</sup>(rate) dt, recognizing this pattern quickly saves time.</li>

</ul>

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

<h3>How do I know whether to integrate with respect to x or y?</h3>

<p>Look at how the boundaries are most naturally described. If the region is bounded by horizontal lines and functions given as y in terms of x, integrate in x. If it's bounded by vertical lines and functions more naturally solved as x in terms of y, integrate in y, this often makes the setup significantly simpler.</p>

</div>

<h3>Do I need to know the shell method for the AP exam?</h3>

<p>No. The AP Calculus AB and BC exams test volumes of revolution using disks and washers, along with cross sections. The shell method is a valid technique and can be a helpful problem-solving tool, but it isn't required knowledge for the exam.</p>

</div>

<h3>What's the difference between distance and displacement, and why does it matter?</h3>

<p>Displacement is where a particle ends up relative to where it started (net change in position). Distance is how much ground it actually covered, including any back-and-forth motion. A particle can return exactly to its starting point (zero displacement) after traveling a large total distance.</p>

</div>

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

<p>The <a href="https://theschoolofmathematics.com/quiz/applications-of-integration-quiz-1">Applications of Integration 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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<a class="cta-btn cta-primary" href="https://theschoolofmathematics.com/quiz/applications-of-integration-quiz-1">Practice Applications of Integration Free</a>

<a class="cta-btn cta-secondary" href="https://theschoolofmathematics.com/quiz/course/AP-Calculus-AB-QBank">Explore the Full AP Calculus AB Qbank</a>

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