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Unit 8 · Applications of Integration Flashcards Cheat Sheet Essentials Visual Review MC Practice FRQ Practice

AP Calculus AB/BC Unit 8 Visual Review

A topic-by-topic visual walkthrough of Unit 8: Applications of Integration — average value, accumulation, area between curves, and volumes of solids.

← Back to Unit 8 hub
TOPIC 8.1 Average Value of a Function f_avg = (1/(b−a)) ∫ₐᵇ f(x) dx Worked example f(x) = x² on [0, 3] f_avg = (1/3)∫₀³ x² dx = (1/3)·(9) = 3 The height of a rectangle on [a,b] with the SAME AREA as under the curve. Don't forget 1/(b−a) The integral alone gives total accumulation; DIVIDE by the interval length for average. By the MVT for integrals, f attains its average value somewhere on [a, b]. Average velocity = displacement / time. f_avg = (1/(b−a))∫ₐᵇ f dx — the integral divided by the interval length. The Review Hub · AP Calculus AB/BC Unit 8 TOPIC 8.2 Motion with Integrals displacement = ∫ₐᵇ v(t) dt position = s(a) + ∫ₐᵗ v dt distance = ∫ₐᵇ |v(t)| dt integration undoes the differentiation from Unit 4 Displacement vs. distance DISPLACEMENT = net change in position (signed — ∫v). DISTANCE = total path length (∫|v|, always ≥ 0). Velocity from acceleration v(t) = v(0) + ∫₀ᵗ a dt Integrate a to get v, then v to get s — each with an initial condition. Split ∫|v| where v changes sign. Displacement = ∫v dt; distance = ∫|v| dt. The Review Hub · AP Calculus AB/BC Unit 8 TOPIC 8.3 Accumulation in Applied Contexts final amount = initial amount + ∫ₐᵇ (rate) dt Tank example Water enters at R_in(t) and leaves at R_out(t) gallons/min. V(T) = V(0) + ∫₀ᵀ (R_in − R_out) dt The integral gives the NET change over [0, T]; add the initial volume. Interpret with units The integral of a rate carries the rate's units × time units. gal/min × min = gallons. Maximum amount occurs where the net rate switches + → − (R_in = R_out). final = initial + ∫(rate)dt — the accumulation-function FRQ pattern. The Review Hub · AP Calculus AB/BC Unit 8 TOPIC 8.4 Area Between Curves (in x) top f(x) bottom g(x) A = ∫ₐᵇ [top − bottom] dx The steps 1. Find intersections → limits a, b. 2. Determine which curve is on TOP. 3. Integrate (top − bottom). ∫₀¹ (√x − x²) dx = 1/3 The integrand is always ≥ 0 in the region. Area between curves = ∫(top − bottom) dx between intersection points. The Review Hub · AP Calculus AB/BC Unit 8 TOPIC 8.5 Area Between Curves (in y) A = ∫_c^d [right − left] dy Integrate with respect to y When curves are easier as x = f(y), or a horizontal slice is simpler, integrate in y. Use RIGHT curve minus LEFT curve, and y-limits c to d. Solve each equation for x in terms of y first. Which variable to integrate? Slice PERPENDICULAR to the axis you integrate along. Vertical strips → dx (top − bottom). Horizontal strips → dy (right − left). Pick whichever avoids splitting the region. In y: area = ∫(right − left) dy — use horizontal strips. The Review Hub · AP Calculus AB/BC Unit 8 TOPIC 8.6 Curves Crossing More Than Twice The top curve can switch When two curves cross several times, the "upper" function changes between them. SPLIT the integral at each intersection and use (top − bottom) on each piece. Add the separate areas together. Worked structure Cross at x = a, m, b (m in the middle): A = ∫ₐᵐ (f−g) dx + ∫ᵐᵇ (g−f) dx Each integrand is the higher minus lower on that subinterval. Or use absolute value The total area is ∫ₐᵇ |f − g| dx — but on the exam you still split it into pieces to integrate. Split at each crossing, integrate (top − bottom) on each piece, and add. The Review Hub · AP Calculus AB/BC Unit 8 TOPIC 8.7 Cross Sections: Squares & Rectangles V = ∫ₐᵇ A(x) dx A(x) = area of the cross section Build a solid from slices The base is a region; each slice ⊥ to the x-axis is a square/rectangle. The side length s = (top − bottom) of the base at that x. square: A(x) = s² = (top−bottom)² Worked example Base bounded by y=√x and y=0 on [0,4], square cross sections: V = ∫₀⁴ (√x)² dx = ∫₀⁴ x dx = 8 Rectangle: A = base × given height. V = ∫A(x)dx; for squares A = s² where s is the region's width. The Review Hub · AP Calculus AB/BC Unit 8 TOPIC 8.8 Cross Sections: Triangles & Semicircles // area formulas for the slice, with s = width of base at x equilateral triangle: A = (√3/4)·s² isosceles right (legs s): A = ½s² semicircle (diameter s): A = (π/8)·s² Plug the area formula into V = ∫A(x)dx Find s = (top − bottom) from the base region, substitute into the shape's area formula, then integrate. e.g. semicircles on base y=√x, [0,4]: V = ∫₀⁴ (π/8)(√x)² dx = (π/8)·8 = π For a semicircle, radius = s/2, so area = ½π(s/2)² = (π/8)s². Use the shape's area formula in V=∫A(x)dx (triangle, semicircle). The Review Hub · AP Calculus AB/BC Unit 8 TOPIC 8.9 Disc Method (x- or y-axis) V = π ∫ₐᵇ [R(x)]² dx Solid of revolution Rotate a region touching the axis → each slice is a solid DISC of radius R. R(x) = distance from the axis to the curve. y=√x about x-axis, [0,4]: V = π∫₀⁴ (√x)² dx = 8π About the y-axis Integrate in y with radius R(y): V = π ∫_c^d [R(y)]² dy Solve the curve for x = f(y) to get R(y). Use discs when the region TOUCHES the axis of rotation (no gap). Disc method: V = π∫[R]² dx when the region touches the axis. The Review Hub · AP Calculus AB/BC Unit 8 TOPIC 8.10 Disc Method: Other Axes rotate about y = k: R(x) = | curve − k | Adjust the radius for the new axis The radius is the DISTANCE from the axis of rotation to the curve — not just the curve's y-value. about y=−1: R = f(x) − (−1) = f(x)+1 about x=a for y-rotation: R = a − x. Worked example y=√x, [0,4], rotate about y = −1: V = π∫₀⁴ (√x + 1)² dx Everything else is the same — only the radius expression changes. Sketch it to get the radius right. For a shifted axis, the radius is the distance to that axis (e.g. curve + 1). The Review Hub · AP Calculus AB/BC Unit 8 TOPIC 8.11 Washer Method (x- or y-axis) V = π ∫ₐᵇ ( [R_outer]² − [R_inner]² ) dx A disc with a hole When the region does NOT touch the axis, each slice is a WASHER (ring). R_outer = far curve to the axis; R_inner = near curve to the axis. Subtract the SQUARES, not (R_out − R_in)². Worked example Region between y=x and y=x², [0,1], rotated about the x-axis: V = π∫₀¹ (x² − (x²)²) dx = π∫₀¹ (x² − x⁴) dx = 2π/15 Outer = x (larger), inner = x² on [0,1]. Washer: V = π∫(R_out² − R_in²)dx when there's a gap to the axis. The Review Hub · AP Calculus AB/BC Unit 8 TOPIC 8.12 Washer Method: Other Axes about y=k: R = |curve − k|, both outer & inner Shift BOTH radii Measure both R_outer and R_inner from the shifted axis. about y=3: R = 3 − curve The curve FARTHER from the axis is R_out; the nearer one is R_in. A reliable checklist 1. Sketch the region and the axis. 2. Write each radius as a distance. 3. Decide disc (touches) vs washer (gap). 4. Integrate π∫(R_out² − R_in²). Rotating about a line inside the region needs care. Measure both radii from the shifted axis; farther curve = outer. The Review Hub · AP Calculus AB/BC Unit 8 TOPIC 8.13 BC ONLY Arc Length L = ∫ₐᵇ √(1 + [f′(x)]²) dx Length of a curve Sum tiny hypotenuses √(dx² + dy²) along the curve → the arc length integral. Compute f′, square it, add 1, take the root, integrate. Often evaluated numerically on the exam. Distance traveled (parametric) For a particle x(t), y(t): L = ∫ √((dx/dt)² + (dy/dt)²) dt This is the path length (speed integrated over time) — see Unit 9. speed = √((x′)² + (y′)²). Arc length (BC): L = ∫√(1 + (f′)²) dx. The Review Hub · AP Calculus AB/BC Unit 8
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How to use the visual review

Spend 30 seconds per slide before clicking next. Look at the diagram, then ask yourself: "Could I set up this integral from the picture?"

Use the fullscreen button () on desktop for the best experience. Use arrow keys to navigate. Tap "Show all slides" to jump around.

This is great for review the night before the exam — fast, visual, and covers every idea you need to recognize in Unit 8.