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Flashcards · AP® Calculus AB · Unit 6

Unit 6: Integration and Accumulation of Change flashcards

73 cards on topics 6.1 to 6.14. Pick a topic or study the whole unit.

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6.1 Exploring Accumulations of Change

What does the area under a rate-of-change graph represent?
The change in the quantity over that interval. Its units are rate units × input units, e.g. (L/min) × min = L. It is a change, not an amount.
What is the difference between net change and total change from a rate graph?
Net change = area above the axis − area below. Total change (e.g. distance) = area above + area below.
A rate graph has area 7 above the axis and 10 below on [0, 9], and the quantity starts at 25. Find the amount at t = 9 and the total change.
Amount: 25 + 7 − 10 = 22. Total change: 7 + 10 = 17.
How do you find the amount at the end from a rate graph?
Amount at the end = amount at the start + net change. Giving the change alone as the final amount is a common mistake.
Which geometric areas are useful for rate graphs made of segments and arcs?
Rectangle bh; triangle ½bh; trapezoid ½(a + b)h; semicircle ½πr²; quarter circle ¼πr². Use them only when the pieces really are lines or circular arcs.
Oil leaks at a constant 0.6 litres per hour for 5 hours. How much leaks out?
Change = rate × time = 0.6 × 5 = 3 litres, the area of a rectangle.
How do you find the average rate over [a, b] from a rate graph?
Net change ÷ (b − a). For example, 2,450 passengers over 20 minutes is 122.5 passengers per minute.

6.2 Approximating Areas with Riemann Sums

How are left, right, midpoint and trapezoidal sums formed on one subinterval?
Left: f(left end) × width. Right: f(right end) × width. Midpoint: f(middle) × width. Trapezoid: ½[f(left) + f(right)] × width.
How are the trapezoidal sum and left and right sums related?
On the same partition, T = (L + R)/2.
When are left and right sums over- or underestimates?
If f is increasing on the interval: L is too small, R too big. If f is decreasing: R is too small, L too big. Monotonicity decides L and R.
When are midpoint and trapezoidal sums over- or underestimates?
If f is concave up: M is too small, T too big. If f is concave down: T is too small, M too big. Concavity decides M and T; if f is linear both are exact.
For f(x) = x² on [1, 3] with two subintervals of width 1, find L, R, T and M.
L = 1 + 4 = 5; R = 4 + 9 = 13; T = 9; M = 1.5² + 2.5² = 2.25 + 6.25 = 8.5. The exact area is 26/3 ≈ 8.667.
What mistakes are common with Riemann sums from tables?
Using Δx = (b − a)/n when widths are unequal; including the last value in a left sum; averaging two values and calling it a midpoint value; forgetting the ½ in trapezoids.
When can you use a midpoint sum from a table?
Only if the table gives values at the midpoints of the subintervals you use.

6.3 Riemann Sums, Summation Notation, and Definite Integral Notation

How is the definite integral defined from Riemann sums?
For f continuous on [a, b], ∫ (a to b) f(x) dx = lim (n → ∞) Σ (i = 1 to n) f(a + iΔx)Δx with Δx = (b − a)/n. Any choice of sample points gives the same limit.
Evaluate Σ (i = 1 to 4) (2i + 1).
3 + 5 + 7 + 9 = 24.
Write lim (n → ∞) Σ (i = 1 to n) (1 + 2i/n)²·(2/n) as an integral.
The width 2/n gives b − a = 2. It is ∫ (1 to 3) x² dx, or equivalently ∫ (0 to 2) (1 + x)² dx.
In a limit of Riemann sums, what does the number in (number)·i/n tell you?
It is the width b − a, not b. Δx becomes dx and a + iΔx becomes x.
Write ∫ (0 to 2) 3x dx as a limit of a right Riemann sum and find its value.
lim (n → ∞) Σ (i = 1 to n) 3(2i/n)(2/n). Geometrically it is a triangle with base 2 and height 6, so the value is 6.
Why is a definite integral called a signed area?
Parts of the region below the axis count as negative.

6.4 The Fundamental Theorem of Calculus and Accumulation Functions

What does the first Fundamental Theorem of Calculus say about g(x) = ∫ (a to x) f(t) dt?
If f is continuous on an interval containing a, then g′(x) = f(x). So every continuous function has an antiderivative.
How do you differentiate ∫ (a to u(x)) f(t) dt?
f(u(x))·u′(x), by the chain rule. With x as the lower limit, d/dx ∫ (x to b) f(t) dt = −f(x).
Find d/dx ∫ (0 to 5x) cos(t²) dt.
cos((5x)²)·5 = 5 cos(25x²).
Find d/dx ∫ (x to 4) sin(√t) dt.
−sin(√x), for x > 0. The minus sign comes from x being the lower limit.
If g(x) = ∫ (3 to x) f(t) dt, what is g(3)?
0, because the interval has zero width. It is not f(3).
What mistakes are common with accumulation functions?
Differentiating the integrand instead of substituting x into it; writing g(a) = f(a); missing u′(x); missing the minus when x is the lower limit; writing x inside the integral as well as in a limit.

6.5 Interpreting the Behavior of Accumulation Functions Involving Area

If g(x) = ∫ (a to x) f(t) dt and you are given the graph of f, what tells you about g?
Heights of f are slopes of g; areas under f are changes in g; slopes of f give the concavity of g. Values of g come from signed areas starting at g(a) = 0.
Where does g(x) = ∫ (a to x) f(t) dt have relative extrema?
Where f changes sign: + to − gives a relative maximum of g; − to + gives a relative minimum. A zero of f with no sign change gives no extremum.
f has a relative maximum at t = 2. What does g(x) = ∫ (a to x) f(t) dt have at x = 2?
A point of inflection, because g″ = f′ changes from + to −.
g(x) = ∫ (2 to x) f(t) dt and ∫ (0 to 2) f(t) dt = 5. Find g(0).
g(0) = ∫ (2 to 0) f(t) dt = −5. Areas to the left of the start count with the opposite sign.
How do you find the absolute extrema of g on [p, q]?
List the endpoints and the zeros of f with a sign change, find g at each from areas, and compare.
Why is "f is largest at c, so g is largest at c" wrong?
f largest means g is rising fastest there. Extrema of g come from sign changes of f, not from where f is biggest.

6.6 Applying Properties of Definite Integrals

What are the main properties of definite integrals?
∫ (a to a) f = 0; ∫ (b to a) f = −∫ (a to b) f; ∫ k·f = k∫ f; ∫ (f ± g) = ∫ f ± ∫ g; ∫ (a to b) f + ∫ (b to c) f = ∫ (a to c) f; ∫ (a to b) k dx = k(b − a).
∫ (1 to 9) f = 4 and ∫ (5 to 9) f = 6. Find ∫ (1 to 5) f.
∫ (1 to 5) f = 4 − 6 = −2.
∫ (2 to 5) f = 7. Find ∫ (5 to 2) [f(x) + 1] dx.
Reversing the limits: −(7 + 1 × 3) = −10. The constant adds k(b − a), not k.
Find ∫ (−6 to 6) √(36 − x²) dx.
The graph is a semicircle of radius 6 above the axis, so the integral is ½π(6²) = 18π.
Which "rules" for integrals are false?
∫ fg ≠ (∫ f)(∫ g); ∫ |f| ≠ |∫ f| unless f keeps one sign; ∫ (f + k) ≠ ∫ f + k.
How do holes and jumps affect a definite integral?
Changing f at one point does not change the integral. At a jump, split the interval and use the right formula on each piece. Vertical asymptotes are not allowed here.

6.7 The Fundamental Theorem of Calculus and Definite Integrals

What does the second Fundamental Theorem of Calculus state?
If f is continuous on [a, b] and F is any antiderivative of f, then ∫ (a to b) f(x) dx = F(b) − F(a). No + C is needed because it cancels.
How is the Fundamental Theorem used for net change and a final amount?
∫ (a to b) F′(x) dx = F(b) − F(a), so F(b) = F(a) + ∫ (a to b) F′(x) dx.
Evaluate ∫ (1 to 9) √x dx.
[(2/3)x^(3/2)] (1 to 9) = (2/3)(27) − 2/3 = 18 − 2/3 = 52/3.
Evaluate ∫ (0 to π) cos x dx.
[sin x] (0 to π) = 0 − 0 = 0. The areas above and below the axis cancel.
Why can you not use the Fundamental Theorem on ∫ (−1 to 2) 3/x⁴ dx?
The integrand is not continuous on [−1, 2]: it is undefined at x = 0 inside the interval.
What mistakes are common when evaluating with the Fundamental Theorem?
F(a) − F(b) instead of F(b) − F(a); no brackets around F(a); substituting into f instead of F; the antiderivative of sin x is −cos x; calling net change "distance".

6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation

What does ∫ f(x) dx mean?
The indefinite integral: all antiderivatives of f. If F′ = f, then ∫ f(x) dx = F(x) + C. Graphically, one curve shifted up and down.
What are the basic antiderivatives of xⁿ, 1/x, eˣ and aˣ?
∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C for n ≠ −1; ∫ (1/x) dx = ln|x| + C; ∫ eˣ dx = eˣ + C; ∫ aˣ dx = aˣ/ln a + C.
What are the basic trig antiderivatives?
∫ cos x dx = sin x + C; ∫ sin x dx = −cos x + C; ∫ sec²x dx = tan x + C; ∫ csc²x dx = −cot x + C; ∫ sec x tan x dx = sec x + C; ∫ csc x cot x dx = −csc x + C.
Which antiderivatives give inverse trig functions?
∫ 1/√(1 − x²) dx = arcsin x + C and ∫ 1/(1 + x²) dx = arctan x + C.
Find ∫ x^(2/3) dx and ∫ 1/x² dx.
(3/5)x^(5/3) + C; and ∫ x⁻² dx = −x⁻¹ + C = −1/x + C (not a logarithm).
How do you integrate a product such as (x + 1)²?
There is no product rule for integrals: expand, split or use an identity first. ∫ (x² + 2x + 1) dx = x³/3 + x² + x + C.
What mistakes are common with indefinite integrals?
No + C; using the power rule on 1/x; writing ln for every fraction; integrating factor by factor; trig sign slips; missing dx.

6.9 Integrating Using Substitution

What is integration by substitution?
The chain rule backwards: ∫ f(g(x))g′(x) dx = F(g(x)) + C. Choose u = g(x), find du, rewrite everything in u, integrate, substitute back, add + C.
How do you choose u for a substitution?
Look for an inner function whose derivative is present up to a constant: the base of a power, the inside of a root, the exponent of e, the input of sin or cos, or a denominator whose derivative is in the numerator.
Find ∫ 3x²e^(x³) dx.
With u = x³, du = 3x² dx, so the integral is ∫ eᵘ du = e^(x³) + C.
Find ∫ sin(4x) dx.
−(1/4) cos(4x) + C. For a linear inside, ∫ f(ax + b) dx = (1/a)F(ax + b) + C.
Evaluate ∫ (0 to 2) x(x² + 1)² dx.
u = x² + 1 runs from 1 to 5, and x dx = du/2: (1/2)∫ (1 to 5) u² du = (1/2)(125 − 1)/3 = 62/3.
How do you handle the limits in a definite integral by substitution?
Change them to u = g(a) and u = g(b) and finish in u, or return to x and use the original limits. Never mix the two.
What can and cannot be adjusted when du does not match exactly?
Constants can be adjusted (du = 2x dx gives x dx = du/2). Variables cannot: never divide by a variable to make du appear.
What is ∫ tan x dx?
−ln|cos x| + C, by substitution u = cos x, since tan x = sin x/cos x.

6.10 Integrating Functions Using Long Division and Completing the Square

When should you use long division before integrating a rational function?
When the degree of the top is at least the degree of the bottom. Rewrite as quotient + remainder/divisor, and keep the remainder.
Find ∫ x²/(x + 1) dx.
x²/(x + 1) = x − 1 + 1/(x + 1), so the integral is x²/2 − x + ln|x + 1| + C.
What are the arctan and arcsin integration forms?
∫ du/(a² + u²) = (1/a) arctan(u/a) + C and ∫ du/√(a² − u²) = arcsin(u/a) + C, with a > 0 the square root of the constant.
Find ∫ 1/(x² + 2x + 10) dx.
Complete the square: x² + 2x + 10 = (x + 1)² + 9, so a = 3 and the integral is (1/3) arctan((x + 1)/3) + C.
Find ∫ 1/√(3 + 2x − x²) dx.
3 + 2x − x² = 4 − (x − 1)², so a = 2 and the integral is arcsin((x − 1)/2) + C.
How do you complete the square on x² + bx + c?
x² + bx + c = (x + b/2)² + (c − b²/4). It is a sum of squares (giving an arctan) when b² − 4c is negative.
What mistakes are common in these integrals?
Integrating top and bottom separately; dropping the remainder; using a² instead of a (1/4 instead of 1/2 when the constant is 4); writing ln(quadratic) when the top is not a multiple of the bottom's derivative.

6.14 Selecting Techniques for Antidifferentiation

In what order should you consider integration techniques?
Basic rule? Rewrite? Substitution? Long division? Complete the square? Then integration by parts or partial fractions where the course includes them.
What signals that substitution will work?
An inner function g(x) with g′(x) present, up to a constant factor. Example: ∫ x/(x² + 9) dx = ½ ln(x² + 9) + C.
Why is ∫ 1/(x² + 9) dx not a logarithm?
The top is not a multiple of the derivative of the bottom. It is an arctan form: (1/3) arctan(x/3) + C.
Find ∫ (x + 1)/x dx.
Split first: ∫ (1 + 1/x) dx = x + ln|x| + C.
Find ∫ x/(x² + 1)² dx.
u = x² + 1, x dx = du/2: (1/2)∫ u⁻² du = −1/(2(x² + 1)) + C.
What do you do with an integrand such as e^(−x²) or sin(x²)?
It has no antiderivative in familiar functions. Use technology for a definite value, or write an accumulation function.
Find ∫ 1/(x² + 2x + 5) dx.
x² + 2x + 5 = (x + 1)² + 4, so the integral is ½ arctan((x + 1)/2) + C.

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