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Flashcards · AP® Calculus AB · Unit 8

Unit 8: Applications of Integration flashcards

75 cards on topics 8.1 to 8.12. Pick a topic or study the whole unit.

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8.1 Finding the Average Value of a Function on an Interval

What is the average value of a continuous function f on [a, b]?
f_avg = (1/(b − a)) ∫ (a to b) f(x) dx, in the units of f. It is the height of the rectangle on [a, b] with the same signed area as the region under f.
Find the average value of 6x² on [0, 2].
∫ (0 to 2) 6x² dx = [2x³] (0 to 2) = 16, divided by 2 gives 8.
The average value of f on [3, 8] is 7. Find ∫ (3 to 8) f(x) dx.
7 × (8 − 3) = 35. Total accumulation = average value × interval length.
What is the difference between the average value of f and the average rate of change of f?
Average value: (1/(b − a)) ∫ (a to b) f(x) dx, in units of f. Average rate of change: (f(b) − f(a))/(b − a), in units of f per unit x. The average value of f′ equals the average rate of change of f.
Is the average value of f equal to (f(a) + f(b))/2?
Only when f is linear. For 2x + 1 on [0, 4] the average is 5 = (1 + 9)/2, because the region is a trapezoid. In general it is not.
Does f reach its average value on [a, b]?
If f is continuous, yes, at least once, by the Intermediate Value Theorem. That c need not be the midpoint or unique.
How do you estimate an average value from a table?
Estimate the integral with a Riemann or trapezoidal sum (weighting each value by its width), then divide by b − a. Do not simply average entries with unequal gaps.

8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals

What is the difference between displacement and total distance from v(t)?
Displacement = ∫ (a to b) v(t) dt = x(b) − x(a), a signed value. Total distance = ∫ (a to b) |v(t)| dt, never negative.
How do you find position at time b from velocity?
x(b) = x(a) + ∫ (a to b) v(t) dt. Integration gives only the change, so you need the starting position. Similarly v(b) = v(a) + ∫ (a to b) a(t) dt.
v(t) = 4 − 2t on [0, 3]. Find the displacement and the total distance.
v changes sign at t = 2. Displacement ∫ (0 to 3) (4 − 2t) dt = 3. Distance: 4 forward on [0, 2] and 1 back on [2, 3], so 5.
How do you find total distance by hand?
Split at the times where v changes sign, integrate v on each piece, take absolute values and add. With a calculator, integrate |v| directly.
x(0) = 5 and ∫ (0 to 2) v(t) dt = −7. Find x(2).
x(2) = 5 + (−7) = −2.
What is average speed compared with average velocity?
Average velocity = (1/(b − a)) ∫ v dt (displacement over time). Average speed = total distance ÷ (b − a).
What mistakes are common with motion integrals?
Calling ∫ v dt the distance; writing |∫ v dt| (the absolute value goes inside); forgetting x(a); splitting where a = 0 instead of where v changes sign; giving ∫ a dt units of distance.

8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts

What does the integral of a rate represent in context?
The net change in the amount: ∫ (a to b) Q′(t) dt = Q(b) − Q(a), in rate units × time units. The amount at time x is Q(a) + ∫ (a to x) Q′(t) dt.
A tank holds 30 litres at t = 0, and water flows in at 4 + 2t litres per minute. How much is in the tank at t = 5?
30 + ∫ (0 to 5) (4 + 2t) dt = 30 + (20 + 25) = 75 litres.
How do you model an amount with a rate in and a rate out?
Q(x) = Q(a) + ∫ (a to x) [I(t) − O(t)] dt. The amount increases exactly when in > out.
When is an amount greatest, given rates in and out?
Not when the rate peaks. Compare the amount at the endpoints and where the net rate in − out changes sign (from + to − for a maximum).
r(t) is in litres per minute and t in minutes. What are the units of ∫ (0 to 10) r(t) dt?
Litres.
What mistakes are common with accumulation in context?
Leaving out the starting amount; using rate(b) − rate(a) as the change in amount; writing out − in; mixing minutes and hours; rounding a critical value before a later integral; no units.

8.4 Finding the Area Between Curves Expressed as Functions of x

How do you find the area between two curves with vertical strips?
Area = ∫ (a to b) [top − bottom] dx, where the limits are the left and right edges (given lines, or where the curves meet). Test one x-value inside to see which curve is on top.
Find the area enclosed by y = x² and y = x.
They meet at x = 0 and x = 1, with x on top: ∫ (0 to 1) (x − x²) dx = 1/2 − 1/3 = 1/6.
Find the area between y = 4 − x² and the x-axis.
Limits −2 and 2: ∫ (−2 to 2) (4 − x²) dx = 16 − 16/3 = 32/3.
Does top − bottom still work if the region is below the x-axis?
Yes. The x-axis is just the curve y = 0. For f below the axis, the area is ∫ (a to b) [0 − f(x)] dx.
When must an area integral in x be split?
When the top or bottom boundary changes formula at some x = c. Split at c and use the correct top and bottom on each piece.
Your area integral gives −5. What went wrong?
You subtracted bottom − top. The area is 5.

8.5 Finding the Area Between Curves Expressed as Functions of y

How do you find area with horizontal strips?
Area = ∫ (c to d) [R(y) − L(y)] dy, right curve minus left curve, with every boundary written as x in terms of y and y-limits.
How do you rewrite curves for a dy area integral?
Solve for x: y = ln x becomes x = eʸ, and y = x³ becomes x = ∛y. The integrand must contain only y.
Find the area enclosed by x = y² and x = 4.
They meet at y = ±2, with x = 4 on the right: ∫ (−2 to 2) (4 − y²) dy = 16 − 16/3 = 32/3.
When is a dy set-up better than dx?
When it needs one integral instead of several, or avoids ± square roots. Both give the same area.
In a dy set-up, which question replaces "which curve is on top?"
"Which curve is on the right?" Test one y-value between the limits; the larger x is on the right.
Your dy area integral gives −9. What went wrong?
You subtracted left − right. The area is 9.

8.6 Finding the Area Between Curves That Intersect at More Than Two Points

How do you find the area between curves that cross inside the interval?
Find every crossing, make a sign chart for f − g, and add ∫ [top − bottom] dx over each piece. The strip height is |f(x) − g(x)|.
How can a calculator find area between crossing curves in one step?
Enter ∫ (a to b) |f(x) − g(x)| dx. You still need the crossings if they are the ends of the region.
On three pieces, ∫ (f − g) dx equals 4, −7 and 2. Find the area and the net value.
Area 4 + 7 + 2 = 13. Net (signed) value 4 − 7 + 2 = −1.
When is |∫ (a to b) (f − g) dx| the area between the curves?
Only when the curves do not cross between a and b. Otherwise the pieces cancel.
Two curves cross at x = 1 and x = 4 inside [0, 6], and nowhere else. How many integrals do you need without a calculator?
Three: on [0, 1], [1, 4] and [4, 6].
What mistakes are common with crossing curves?
One integral of f − g across all pieces; absolute value outside the integral; missing a crossing (e.g. at x = 0); one test value for the whole interval; splitting at the x-axis instead of at crossings.

8.7 Volumes with Cross Sections: Squares and Rectangles

How do you find the volume of a solid with known cross sections?
V = ∫ (a to b) A(x) dx, where A(x) is the area of the cross section at x. Perpendicular to the x-axis, the side s(x) = top − bottom with dx; perpendicular to the y-axis, s(y) = right − left with dy.
What is the volume when cross sections perpendicular to the x-axis are squares?
V = ∫ (a to b) [s(x)]² dx with s(x) = top − bottom. Find s first, then square it: (f − g)², not f² − g².
The base is under y = √x on [0, 4], and cross sections perpendicular to the x-axis are squares. Find the volume.
A = (√x)² = x, so V = ∫ (0 to 4) x dx = 8.
What are the cross-section areas for rectangles?
Height h(x): A = s·h. Height k times the base: A = ks². Constant height h: V = h × (area of the base).
A base region has area 5, and slices are rectangles of constant height 3. Find the volume.
V = ∫ 3s dx = 3 × 5 = 15.
What mistakes are common with square and rectangle cross sections?
Forgetting to square (∫ s dx is the base area); writing f² − g²; putting π in a square's area; mixing x and y in a dy integral.

8.8 Volumes with Cross Sections: Triangles and Semicircles

What are the areas of triangle and semicircle cross sections on side s?
Equilateral triangle (√3/4)s²; isosceles right triangle with a leg in the base s²/2; with the hypotenuse in the base s²/4; semicircle with diameter in the base (π/8)s².
Why is a semicircle cross section's area (π/8)s²?
The diameter is s, so the radius is s/2 and the area is ½π(s/2)² = πs²/8. Using s as the radius makes the answer four times too big.
A semicircular slice has diameter 6. What is its area?
Radius 3, so ½π(3)² = 9π/2.
An equilateral triangular slice has side 2. What is its area?
(√3/4)(2²) = √3. The area constant is √3/4, not √3/2.
Two solids share a base: one has isosceles right triangles with the hypotenuse in the base, the other with a leg in the base. What is the ratio of volumes?
1 : 2, since (1/4) ÷ (1/2) = 1/2.
How is the volume set up for any standard cross-section shape?
Each has A = c·s², so V = c ∫ (a to b) [s(x)]² dx, with s = top − bottom (or right − left in y).

8.9 Volume with Disc Method: Revolving Around the x- or y-Axis

What is the disc method for revolving about the x-axis?
V = π ∫ (a to b) [f(x)]² dx, where the radius f(x) is the distance from the axis to the curve. The region must touch the axis along the whole interval.
How do you use the disc method about the y-axis?
Use horizontal slices: write the curve as x = g(y) and compute V = π ∫ (c to d) [g(y)]² dy with y-limits.
The region under y = √x for 0 ≤ x ≤ 4 is revolved about the x-axis. Find the volume.
π ∫ (0 to 4) (√x)² dx = π ∫ (0 to 4) x dx = 8π.
The region between y = x² (x ≥ 0), the y-axis and y = 4 is revolved about the y-axis. Find the volume.
x = √y, so V = π ∫ (0 to 4) y dy = 8π.
Is π ∫ (0 to 2) x³ dx the volume when the region under y = x³ on [0, 2] is revolved about the x-axis?
No. The radius x³ must be squared: V = π ∫ (0 to 2) x⁶ dx = 128π/7.
What mistakes are common with the disc method?
No π or 2π instead of π; not squaring the radius; dx with a y-axis rotation; using the diameter as the radius; square units on a volume.
What quick check bounds a disc-method volume?
With widest radius R and length L, the volume is less than the cylinder πR²L.

8.10 Volume with Disc Method: Revolving Around Other Axes

How do you use the disc method about a line y = k?
The radius is the distance from the axis to the curve: k − f(x) if the curve is below, f(x) − k if above. V = π ∫ (a to b) [radius]² dx.
What is the radius when y = f(x) is above the axis y = −3?
f(x) − (−3) = f(x) + 3.
The triangle bounded by y = x, y = 2 and the y-axis is revolved about y = 2. Find the volume.
Radius 2 − x, so V = π ∫ (0 to 2) (2 − x)² dx = 8π/3, the volume of a cone with radius 2 and height 2.
A region is revolved about x = 4. Which differential do you use?
dy, because the axis is vertical. The radius is h − g(y) or g(y) − h with x = g(y).
What mistakes are common when revolving about other axes?
Measuring the radius from the x- or y-axis; sign slips with a negative axis; squaring term by term ((1 − √x)² = 1 − 2√x + x); using discs when the region does not reach the axis.
When can the disc method be used about a line?
Only when the region touches that axis along its whole length. If there is a gap, the slices are washers.

8.11 Volume with Washer Method: Revolving Around the x- or y-Axis

When do you use washers instead of discs?
When there is a gap between the region and the axis. Each slice spins into a ring with area π(R² − r²), where R is the distance to the farther edge and r to the nearer edge.
What is the washer method formula about the x-axis?
V = π ∫ (a to b) [R(x)² − r(x)²] dx. For a region above the x-axis, R is the top curve and r is the bottom curve.
A washer has R = 3 and r = 1. What is the area of its face?
π(9 − 1) = 8π.
Why is π(R − r)² wrong for a washer?
The face is a big circle minus a small one: π(R² − r²). With R = 5 and r = 3, R² − r² = 16 but (R − r)² = 4.
The rectangle between y = 1 and y = 3 for 0 ≤ x ≤ 2 is revolved about the x-axis. Find the volume.
R = 3, r = 1: π ∫ (0 to 2) (9 − 1) dx = 16π.
How is the washer method set up about the y-axis?
Horizontal slices: V = π ∫ (c to d) [R(y)² − r(y)²] dy, with R the right curve and r the left curve, both as x in terms of y.

8.12 Volume with Washer Method: Revolving Around Other Axes

How do you find washer radii about a line y = k or x = h?
Each radius is a distance from the axis: larger coordinate minus smaller. R belongs to the boundary farther from the axis, r to the nearer one. Decide with a test point.
A vertical strip runs from y = 1 to y = 3 and is revolved about y = 4. Find R, r and the face area.
R = 4 − 1 = 3, r = 4 − 3 = 1, so the face area is π(9 − 1) = 8π.
If the axis is above the region, is the top curve R?
No. When the axis is above, the lower curve is farther away, so it gives R.
How far is the curve point at y = 2 from the axis y = 6, and from the axis y = −3?
6 − 2 = 4 and 2 − (−3) = 5.
How far is the point x = 3 from the axis x = −2?
3 − (−2) = 5.
What mistakes are common with washers about other axes?
Calling the top curve R when the axis is above; subtracting k for an axis like y = −1 (use f(x) + 1); shifting the final answer instead of each radius; (R − r)² instead of R² − r².

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