True or False? An accumulation of change tells you the total value of the quantity.

False. It tells you only how much the quantity has changed over the interval. To reach a total you also need a boundary value , a known value of the quantity at one particular point.

Define a Riemann sum .

A Riemann sum approximates an accumulation of change, equivalently a definite integral, by adding up the areas of a number of rectangles. Each rectangle's width is one subinterval, and its height is a value of the function taken on that subinterval.

When is a left Riemann sum an underestimate?

When the function is increasing over the whole interval. Each rectangle is then topped at the lowest point of its own subinterval, so every one of them falls short of the region beneath the curve.

True or False? You can tell straight away whether a midpoint Riemann sum is an over or an underestimate.

False. Each rectangle usually overshoots the curve on one side of its midpoint and falls short on the other, so the two errors partly cancel in a way that is not obvious. The left and right sums do have a simple rule, but the midpoint sum does not.

When is a right Riemann sum an underestimate?

When the function is decreasing over the whole interval. Each rectangle is then topped at the lowest point of its own subinterval, which for a decreasing function is the right-hand end.

Define a trapezoidal sum .

A trapezoidal sum approximates a definite integral by adding up the areas of trapezoids rather than rectangles. Each trapezoid spans one subinterval, with its two parallel sides reaching the curve at the two ends of that subinterval.

When is a trapezoidal sum an overestimate?

When the function is concave up over the whole interval. The straight top of each trapezoid then lies above the curve it joins, so every trapezoid overshoots.

True or False? A trapezoidal sum needs all its subintervals to be the same width.

False. The subintervals may have different widths, and each trapezoid simply uses its own. Equal widths matter only because they allow the sum to be written as a single tidier formula.

How is a trapezoidal sum related to the left and right Riemann sums?

It is exactly their average . Each trapezoid's area is the mean of the corresponding left and right rectangles, so averaging the two sums gives the trapezoidal sum and saves remembering a third formula.

True or False? A definite integral with a variable upper limit defines a new function.

True. With fixed limits a definite integral is a single number, but with a variable upper limit its value changes as that limit changes. That is exactly what an accumulation function is.

How do you evaluate a definite integral of a piecewise-defined function?

Integrate each piece separately over its own part of the interval, then add the results. The function need not be continuous at the joins, provided any discontinuity there is removable or a jump rather than essential.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 6: Integration & Accumulation of ChangeRiemann Sums & Definite Integrals

Riemann Sums & Definite Integrals (College Board AP® Calculus BC): Flashcards

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FrontAccumulation of Change
Define an accumulation of change.

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Cards in this collection (54)

Define an accumulation of change.
It is the actual change a quantity undergoes over an interval, worked out from its rate of change.
A harvester gathering 0.5 kilograms of strawberries per meter accumulates $0.5 \times 6 = 3$ kilograms over 6 meters.
Where on the graph of a rate of change function is the accumulation of change?
It is the area between that graph and the $x$ -axis over the interval concerned.
Simple geometry finds it when the region is made of rectangles, triangles or semicircles, and a definite integral finds it otherwise.
True or False?
An accumulation of change tells you the total value of the quantity.
False.
It tells you only how much the quantity has changed over the interval.
To reach a total you also need a boundary value, a known value of the quantity at one particular point.
A rate of change graph encloses a 5 by 2 rectangle and a right triangle of base 3 and height 5, both above the axis. What is the accumulation of change?
It is $17.5$ , in the units of the accumulated quantity.
The rectangle contributes $5 \times 2 = 10$ and the triangle contributes $\frac{1}{2} \times 5 \times 3 = 7.5$ .
Fill in the missing unit in this accumulation of change:
a rate of change measured in meters per second squared, accumulated over an interval measured in seconds, gives an accumulation of change in $______$ per second
The completed statement is: a rate of change measured in meters per second squared, accumulated over an interval measured in seconds, gives an accumulation of change in meters per second.
The units are always the units of the rate of change multiplied by the units of the independent variable.
What does a negative accumulation of change tell you?
That the quantity decreased over the interval.
The rate of change function was negative there, so its graph lay below the $x$ -axis.
A rate of change graph dips below the axis and encloses a semicircle of radius 2 there. What does that region contribute?
It contributes $- 2 π$ , since the semicircle has area $\frac{1}{2} π {(2)}^{2} = 2 π$ and lies below the axis.
A region below the axis always subtracts from the accumulation of change.
A tank holds 7 gallons, and the accumulation of change over the next interval is $17.5 - 2 π$ gallons. How much is in the tank at the end?
About $18.217$ gallons.
Add the boundary value to the accumulation of change, giving $7 + 17.5 - 2 π = 24.5 - 2 π$ .
Define a Riemann sum.
A Riemann sum approximates an accumulation of change, equivalently a definite integral, by adding up the areas of a number of rectangles.
Each rectangle's width is one subinterval, and its height is a value of the function taken on that subinterval.
Fill in the three missing positions used for the rectangle heights in the Riemann sums:
a left Riemann sum takes the value of $f$ at the $______$ side of each rectangle, a right Riemann sum takes it at the $______$ side, and a midpoint Riemann sum takes it at the $______$ of the subinterval
The completed rule is: a left Riemann sum takes the value of $f$ at the left-hand side of each rectangle, a right Riemann sum takes it at the right-hand side, and a midpoint Riemann sum takes it at the midpoint of the subinterval.
All three use the same rectangle widths, namely $x_{i} - x_{i - 1}$ .
When is a left Riemann sum an underestimate?
When the function is increasing over the whole interval.
Each rectangle is then topped at the lowest point of its own subinterval, so every one of them falls short of the region beneath the curve.
True or False?
You can tell straight away whether a midpoint Riemann sum is an over or an underestimate.
False.
Each rectangle usually overshoots the curve on one side of its midpoint and falls short on the other, so the two errors partly cancel in a way that is not obvious.
The left and right sums do have a simple rule, but the midpoint sum does not.
When is a right Riemann sum an underestimate?
When the function is decreasing over the whole interval.
Each rectangle is then topped at the lowest point of its own subinterval, which for a decreasing function is the right-hand end.
A function has values $2.6, 4.8, 12.2$ and $0.7$ at the left ends of subintervals of widths 3, 4, 3 and 2. Find the left Riemann sum.
It is $65.0$ .
The four rectangles give $3 (2.6) + 4 (4.8) + 3 (12.2) + 2 (0.7)$ , each width multiplied by the value at its left-hand end.
In a midpoint Riemann sum, do you take the midpoint of the $x$ values or the midpoint of the function values?
Of the $x$ values.
Find the midpoint of the subinterval first and evaluate $f$ there, so the height is $f (\frac{x_{i - 1} + x_{i}}{2})$ and not an average of two function values.
A function has values $4.8, 12.2, 0.7$ and $- 1.3$ at the right ends of subintervals of widths 3, 4, 3 and 2. Find the right Riemann sum.
It is $62.7$ .
The four rectangles give $3 (4.8) + 4 (12.2) + 3 (0.7) + 2 (- 1.3)$ , and the last one contributes a negative amount.
How can a Riemann sum be made more accurate?
By increasing the number of rectangles $n$ , so that the subintervals become narrower.
The rectangles then follow the curve more closely and their total area is a better approximation.
Define a trapezoidal sum.
A trapezoidal sum approximates a definite integral by adding up the areas of trapezoids rather than rectangles.
Each trapezoid spans one subinterval, with its two parallel sides reaching the curve at the two ends of that subinterval.
Fill in the two missing function values in the area of the $i$ th trapezoid:
$(x_{i} - x_{i - 1}) \cdot \frac{______+______}{2}$ for the trapezoid on that subinterval
The completed area is $(x_{i} - x_{i - 1}) \cdot \frac{f (x_{i - 1}) + f (x_{i})}{2}$ .
That is the width times the average of the two parallel sides, which is the ordinary area formula for a trapezoid.
When is a trapezoidal sum an overestimate?
When the function is concave up over the whole interval.
The straight top of each trapezoid then lies above the curve it joins, so every trapezoid overshoots.
True or False?
A trapezoidal sum needs all its subintervals to be the same width.
False.
The subintervals may have different widths, and each trapezoid simply uses its own.
Equal widths matter only because they allow the sum to be written as a single tidier formula.
How is a trapezoidal sum related to the left and right Riemann sums?
It is exactly their average.
Each trapezoid's area is the mean of the corresponding left and right rectangles, so averaging the two sums gives the trapezoidal sum and saves remembering a third formula.
Subintervals of widths 3, 4, 3 and 2 have endpoint values $2.6, 4.8, 12.2, 0.7$ and $- 1.3$ in order. Find the trapezoidal sum.
It is $63.85$ .
The four trapezoids give $3 \cdot \frac{2.6 + 4.8}{2} + 4 \cdot \frac{4.8 + 12.2}{2} + 3 \cdot \frac{12.2 + 0.7}{2} + 2 \cdot \frac{0.7 - 1.3}{2}$ .
Define an accumulation function.
An accumulation function gives the accumulation of change measured from one chosen starting point up to an endpoint that is allowed to move.
It is most often written as a definite integral, $g (x) = \int_{a}^{x} f (t) d t$ .
Fill in the missing letter in the rule about the dummy variable in $g (x) = \int_{a}^{x} f (t) d t$ :
any letter may be used for the dummy variable inside the integral except $______$ itself
The completed rule is: any letter may be used for the dummy variable inside the integral except $x$ itself.
Using $x$ for both the upper limit and the variable of integration would confuse two quite different roles.
True or False?
A definite integral with a variable upper limit defines a new function.
True.
With fixed limits a definite integral is a single number, but with a variable upper limit its value changes as that limit changes.
That is exactly what an accumulation function is.
A harvester gathers $0.5$ kilograms of strawberries per meter. What is the accumulation function?
It is $0.5 x$ , where $x$ is the distance in meters from the starting point.
Substituting a value for $x$ gives the amount gathered up to that point, and no integral is needed because the rate is constant.
Find $g (x) = \int_{0}^{x} (2 t + \sin t) d t$ .
It is $g (x) = x^{2} - \cos x + 1$ .
The antiderivative is $t^{2} - \cos t$ , and evaluating it between $0$ and $x$ gives $x^{2} - \cos x - (0 - 1)$ .
If $f$ is the rate of change of a quantity $r$ , how is $r (x)$ written using an accumulation function?
As $r (x) = r (0) + \int_{0}^{x} f (t) d t$ .
The integral supplies the change since $x = 0$ , and the known value $r (0)$ supplies the level it started from.
Fill in the two missing limits in the first fundamental theorem of calculus, where $F$ is an antiderivative of $f$ :
$\int_{a}^{b} f (x) d x = F (______) - F (______)$ whenever $f$ is continuous on that interval
The completed theorem is $\int_{a}^{b} f (x) d x = F (b) - F (a)$ .
Note the order: the value at the upper limit comes first, and the value at the lower limit is subtracted from it.
What conditions does the first fundamental theorem of calculus require?
That $f$ is continuous on the closed interval $[a, b]$ , and that $F$ is an antiderivative of $f$ on that interval.
With both in place a definite integral can be found without any limiting process at all.
How do you find $f (x)$ from $f^{'}$ and one known value $f (a)$ ?
Use $f (x) = f (a) + \int_{a}^{x} f^{'} (t) d t$ .
This is the first fundamental theorem rearranged, and it avoids having to find a constant of integration at all.
$f$ satisfies $f^{'} (x) = {(\ln x)}^{2}$ and $f (1) = 2$ . Find $f (5)$ .
It is $6.857$ to three decimal places.
The fundamental theorem gives $f (5) = f (1) + \int_{1}^{5} {(\ln x)}^{2} d x = 2 + 4.857$ .
Fill in the missing function in the second fundamental theorem of calculus:
$\frac{d}{d x} (\int_{a}^{x} f (t) d t) =______$ whenever $f$ is continuous on an interval containing $a$
The completed theorem is $\frac{d}{d x} (\int_{a}^{x} f (t) d t) = f (x)$ .
Differentiating an accumulation function simply returns the function that was being accumulated.
True or False?
$\int_{a}^{x} f (t) d t$ is an antiderivative of $f$ .
True.
The second fundamental theorem of calculus says exactly that, provided $f$ is continuous on an interval containing $a$ .
So a definite integral with a variable upper limit can stand in wherever an antiderivative is wanted.
What is $\frac{d}{d x} (\int_{a}^{g (x)} f (t) d t)$ ?
It is $f (g (x)) \cdot g^{'} (x)$ .
The chain rule supplies the extra factor, because the upper limit is itself a function of $x$ rather than $x$ alone.
Find $\frac{d}{d x} (\int_{a}^{x^{2}} \sin t d t)$ .
It is $2 x \sin (x^{2})$ .
Putting the upper limit in place of $t$ gives $\sin (x^{2})$ , and differentiating that upper limit contributes the factor $2 x$ .
Define a definite integral.
A definite integral $\int_{a}^{b} f (x) d x$ is an operation that outputs a number, fixed by the function being integrated and by the two integration limits $a$ and $b$ .
The function being integrated is called the integrand.
What are the two main interpretations of a definite integral?
If $f$ is a rate of change, it is the accumulation of change as $x$ goes from $a$ to $b$ .
On a graph where $f (x) \geq 0$ across $[a, b]$ , it is the area between the curve and the $x$ -axis over that interval.
Fill in the missing limit value beneath the limit sign, and the missing width factor at the end:
$\int_{a}^{b} f (x) d x = \lim_{n \to______} \sum_{i = 1}^{n} f (x_{i}^{*})______$ where $x_{i}^{*}$ lies in the $i$ th subinterval
The completed definition is $\int_{a}^{b} f (x) d x = \lim_{n \to \infty} \sum_{i = 1}^{n} f (x_{i}^{*}) Δ x_{i}$ .
As the number of subintervals grows without bound, the total area of the approximating rectangles closes in on the exact value.
True or False?
The type of Riemann sum used changes the value of the limit that defines a definite integral.
False.
The point $x_{i}^{*}$ may be any value in the $i$ th subinterval, so left, right and midpoint sums all fit the definition.
As the subintervals shrink towards zero width the choice stops mattering, and every version tends to the same limit.
What does a factor of $\frac{5}{n}$ in a Riemann-sum limit tell you?
That the subintervals all have width $\frac{5}{n}$ , so the whole interval has width 5 and $b - a = 5$ .
With the starting value read off from inside $f$ , that identifies both limits: $\lim_{n \to \infty} \sum_{i = 1}^{n} ({(- 3 + \frac{5}{n} i)}^{2} + 1) \frac{5}{n}$ is $\int_{- 3}^{2} (x^{2} + 1) d x$ .
What are the sum and constant multiple rules for definite integrals?
They are $\int_{a}^{b} (f (x) \pm g (x)) d x = \int_{a}^{b} f (x) d x \pm \int_{a}^{b} g (x) d x$ and $\int_{a}^{b} k f (x) d x = k \int_{a}^{b} f (x) d x$ .
They are the same rules as for indefinite integrals, with the limits carried along unchanged.
Fill in the three missing results about the limits of a definite integral:
$\int_{a}^{a} f (x) d x$ equals $______$ for any $f$ , reversing the two limits multiplies the value by $______$ instead, and $\int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$ equals the integral from $a$ to $______$ overall
The completed results are: $\int_{a}^{a} f (x) d x$ equals 0 for any $f$ , reversing the two limits multiplies the value by $- 1$ instead, and $\int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$ equals the integral from $a$ to $b$ overall.
The last of the three holds for any $c$ lying between $a$ and $b$ .
Given $\int_{2}^{5} f (x) d x = - 3$ and $\int_{2}^{7} f (x) d x = 12$ , find $\int_{5}^{7} f (x) d x$ .
It is $15$ .
The adjacent-interval property gives $\int_{2}^{7} f (x) d x = \int_{2}^{5} f (x) d x + \int_{5}^{7} f (x) d x$ , so $12 = - 3 + 15$ .
Given $\int_{5}^{7} f (x) d x = 15$ and $\int_{5}^{7} g (x) d x = 2$ , find $\int_{7}^{5} (f (x) - 3 g (x)) d x$ .
It is $- 9$ .
Splitting the integrand gives $\int_{7}^{5} f (x) d x - 3 \int_{7}^{5} g (x) d x$ , and reversing both sets of limits turns that into $- 15 + 3 (2)$ .
Define what it means to evaluate a definite integral.
To evaluate a definite integral is to find the single number it is equal to.
In practice you find an antiderivative of the integrand and then substitute the two limits into it, written as ${[F (x)]}_{a}^{b}$ .
True or False?
A definite integral and an indefinite integral both evaluate to a function of $x$ .
False.
A definite integral evaluates to a number, because its limits fix a single value.
An indefinite integral is another function of $x$ .
Why is the constant of integration not needed in a definite integral?
Because it cancels: $(F (b) + C) - (F (a) + C) = F (b) - F (a)$ .
Any antiderivative may therefore be used, so the simplest one is the obvious choice.
Fill in the missing word about the sign of a definite integral:
if $f (x) < 0$ right across $[a, b]$ then $\int_{a}^{b} f (x) d x$ is $______$ as well
The completed statement is: if $f (x) < 0$ right across $[a, b]$ then $\int_{a}^{b} f (x) d x$ is negative as well.
The same holds with the signs reversed, so an integrand that is positive throughout gives a positive integral.
Evaluate $\int_{- \frac{π}{3}}^{\frac{π}{4}} \cos x d x$ .
It is $\frac{\sqrt{2} + \sqrt{3}}{2}$ .
Since $\int \cos x d x = \sin x + C$ , the value is $\sin \frac{π}{4} - \sin (- \frac{π}{3}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}$ .
What is the value of a definite integral when $f$ is positive on part of $[a, b]$ and negative on the rest?
The negative parts subtract from the positive parts, so the total may be positive, negative, or zero.
A definite integral over such an interval is therefore not the same thing as the total area involved.
Why is $\int_{0}^{3} t^{2} (2 - t) d t$ smaller than $\int_{0}^{2} t^{2} (2 - t) d t$ ?
Because $t^{2} (2 - t)$ is negative for $t > 2$ , so the stretch from $2$ to $3$ subtracts from what has already accumulated.
The two values are $- \frac{9}{4}$ and $\frac{4}{3}$ respectively.
How do you evaluate a definite integral of a piecewise-defined function?
Integrate each piece separately over its own part of the interval, then add the results.
The function need not be continuous at the joins, provided any discontinuity there is removable or a jump rather than essential.

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