MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: Riemann sums and the definite integral (MATH 141)

Q: How do I evaluate a definite integral using the definition with Riemann sums?

Write the width of a strip, b minus a over n, and the right endpoint a plus i times that width. Put the endpoint into the function and expand in powers of i. Pull the constants out of the sum, close it with the formulas for the sums of 1, i, i squared and i cubed, and finally let n go to infinity. No antiderivative is used at any step.

Q: Is the left Riemann sum an overestimate or an underestimate?

It depends on the function, not on the word left. If the function increases on the interval, the left sum is an underestimate and the right sum an overestimate. If it decreases, it is the reverse. If the function turns inside the interval, neither sum has a fixed side: split at the turning point and bound each piece separately.

Q: Why can a definite integral be zero or negative when an area cannot?

Because a Riemann sum multiplies each width by the height of the function, with its sign. Rectangles below the horizontal axis are subtracted, so the integral is the area above the axis minus the area below. To find the area of a region, split the interval where the graph crosses the axis and add the pieces as positive numbers.

Q: How do I recognize a limit of sums as a definite integral?

Look for the factor that behaves like a constant over n: it is the width of a strip, and the constant is b minus a. Inside the function, the expression a plus i times that width is the sample point, which gives the left end a. The right end is a plus n times the width. Several readings may work, and they all give the same value.

Q: Is the midpoint sum the average of the left and right sums?

No. The midpoint sum evaluates the function at the centre of each strip, which are new points. The average of the left and right sums is a different estimate. On the function 16 minus x squared from 0 to 4 with four strips, the midpoint sum is 43 while the average of the left and right sums is 42.

This sheet is not a summary of sections 5.1 and 5.2 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on Riemann sums and the definition of the integral in MATH 141 at McGill University, and which precise gesture avoids each loss.

Everything here is done as on the exam, by hand and without an antiderivative: when a question says use the definition, the limit of the sum is the only answer that earns the method marks. Every number quoted below has been checked.

The thread of the chapter

An integral is the limit of a sum of SIGNED rectangles: before any number, write $\Delta x = \frac{b - a}{n}$ and $x_i = a + i\,\Delta x$ , then let the picture decide the sign of each piece and the side, over or under, on which each sum falls.

This chapter is part of MATH 141, Calculus 2 (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Riemann sum
Sigma notation
Sums of powers of integers
Left and right endpoint sums
Midpoint sum
Definite integral
Signed area
Properties of the definite integral
Comparison properties of the integral
Monotonic function
Area of a plane region

The essentials

Three ingredients, written before any number

• The width: $\Delta x = \frac{b - a}{n}$ . The division points: $x_i = a + i\,\Delta x$ for $i = 0, 1, \ldots, n$ . The $a +$ stays even when $a$ is negative.
• The sums: $L_n = \sum_{i=1}^{n} f(x_{i-1})\,\Delta x$ , $R_n = \sum_{i=1}^{n} f(x_i)\,\Delta x$ , $M_n = \sum_{i=1}^{n} f(\bar{x}_i)\,\Delta x$ with $\bar{x}_i = a + \left(i - \frac{1}{2}\right)\Delta x$ . Each has exactly $n$ terms.
• The definition: $\int_a^b f(x)\,dx = \lim_{n \to \infty} R_n$ for $f$ continuous, and $L_n$ , $M_n$ or any other choice of sample points have the same limit.
• The formulas that close the sums, valid from $i = 1$ to $i = n$ only: $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ , $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ , $\sum_{i=1}^{n} i^3 = \left[\frac{n(n+1)}{2}\right]^2$ and $\sum_{i=1}^{n} c = cn$ .
• Monotonicity decides the side: $f$ increasing gives $L_n \le \int_a^b f \le R_n$ , $f$ decreasing gives the reverse, and in both cases $\left|L_n - R_n\right| = \Delta x\left|f(a) - f(b)\right|$ .

The same increasing curve

y = 1 + \frac{x^2}{4}

[0, 4]

: the left rectangles stay under it, the right ones stick out. On a decreasing curve the two pictures swap.

Writing $\Delta x$ and $x_i$ on the first line of the answer, with the $a$ visible, is already a method mark on most MATH 141 papers, and it is the line that prevents the most expensive error of the chapter.

Signed area and the properties you may quote

• An integral is a SIGNED area: area above the axis minus area below. The area of a region is $\int_a^b |f(x)|\,dx$ , computed by splitting at every crossing of the axis.
• $\int_b^a f = -\int_a^b f$ and $\int_a^a f = 0$ : in the definition, $\Delta x = \frac{a - b}{n}$ is then negative or zero.
• $\int_a^b c\,dx = c(b - a)$ , the integral of a constant being a rectangle, never just $c$ .
• Additivity: $\int_a^b f + \int_b^c f = \int_a^c f$ . Linearity: $\int_a^b (\alpha f + \beta g) = \alpha\int_a^b f + \beta\int_a^b g$ .
• Comparison: $f \ge 0$ gives $\int_a^b f \ge 0$ ; $f \ge g$ gives $\int_a^b f \ge \int_a^b g$ ; $m \le f \le M$ gives $m(b - a) \le \int_a^b f \le M(b - a)$ . None of them has a converse.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Over or under? It depends on the direction of f, never on left or right

Read a line as: when $f$ behaves as in the first column on $[a, b]$ , the left and right sums fall on the side given in the next two. The red line is the rule students invent: when $f$ turns inside the interval, no sum has a fixed side.

Behaviour of f on the interval	Left sum	Right sum
increasing	under	over
Example: $x^2$ on $[0, 2]$ , $n = 2$ : $L_2 = 1 < \frac{8}{3} < 5 = R_2$ .
decreasing	over	under
Example: $16 - x^2$ on $[0, 4]$ , $n = 4$ : $L_4 = 50 > \frac{128}{3} > 34 = R_4$ .
constant	exact	exact
Example: $f(x) = 5$ on $[1, 4]$ : $L_n = R_n = 15 = \int_1^4 5\,dx$ for every $n$ .
turns inside	no rule	no rule no fixed side
Example: $x^2$ on $[-1, 2]$ , $n = 3$ : $L_3 = 2 < 3$ , an underestimate. Same form, other result: $x^2$ on $[-2, 1]$ , $n = 3$ : $L_3 = 5 > 3$ , an overestimate, for the same function. What to do: Split at the turning point and bound each monotonic piece separately.

The same table answers a data question: a table of readings gives a guaranteed bound only if the statement says in which direction the quantity moves between readings.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Dropping the left end from the sample point

the whole question: the limit is correct, but of another integral

What not to write

“For $\int_1^3 (x^2 - 2x)\,dx$ : $\Delta x = \frac{2}{n}$ , $x_i = \frac{2i}{n}$ , so the integral is $\frac{8}{3} - 4 = -\frac{4}{3}$ .”

What to write

“ $x_i = 1 + \frac{2i}{n}$ , so $f(x_i) = \frac{4i^2}{n^2} - 1$ and $R_n = \frac{4(n+1)(2n+1)}{3n^2} - 2 \to \frac{2}{3}$ .”

Why: $x_i = \frac{2i}{n}$ runs over $[0, 2]$ , not $[1, 3]$ . The $n = 1$ test catches it at once: $R_1$ must be $(b - a)f(b) = 2f(3) = 6$ , and the wrong formula gives $0$ .

2. Summing a constant once instead of n times

2 marks, and a final answer off by the whole constant term

What not to write

“ $\sum_{i=1}^{n}\left(\frac{4i^2}{n^2} - 1\right)\frac{2}{n} = \frac{8}{n^3}\sum_{i=1}^{n} i^2 - \frac{2}{n} \to \frac{8}{3}$ .”

What to write

“ $\sum_{i=1}^{n} \frac{2}{n} = n \cdot \frac{2}{n} = 2$ , so the limit is $\frac{8}{3} - 2 = \frac{2}{3}$ .”

Why: A term that does not depend on $i$ is still added once per value of $i$ : $\sum_{i=1}^{n} c = cn$ . In a Riemann sum the constant part of $f$ always contributes $c(b - a)$ , the integral of a constant.

3. Keeping the wrong leading term of a sum of squares

2 marks, and every integral of a quadratic comes out wrong

What not to write

“ $\frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} \approx \frac{n \cdot n \cdot n}{6n^3} \to \frac{1}{6}$ .”

What to write

“ $\frac{n(n+1)(2n+1)}{6n^3} = \frac{1}{6} \cdot 1 \cdot \left(1 + \frac{1}{n}\right)\left(2 + \frac{1}{n}\right) \to \frac{2}{6} = \frac{1}{3}$ .”

Why: The factor $2n + 1$ behaves like $2n$ , not like $n$ . Dividing each factor by one power of $n$ before letting $n \to \infty$ makes the $2$ impossible to lose. Check: $\frac{1}{3}$ is the area under $y = x^2$ on $[0, 1]$ , below the triangle's $\frac{1}{2}$ .

4. Reporting an integral of zero as an area of zero

all the marks of an area question

What not to write

“ $\int_{-3}^{1} (x + 1)\,dx = 0$ , so the region between the line and the axis has no area.”

What to write

“The line crosses the axis at $x = -1$ : two triangles of area $2$ , one below and one above. The integral is $-2 + 2 = 0$ , the area is $2 + 2 = 4$ .”

y = x + 1

[-3, 1]

: the triangle below counts

-2

, the one above

+2

. The integral is

0

, the shaded area is

4

Why: A rectangle below the axis enters the sum with a negative height. An integral is a net balance; an area is computed from $|f|$ , after splitting at each crossing.

5. Believing the left sum always underestimates

1 to 2 marks for the justification, and a wrong inequality

What not to write

“ $L_3$ is an underestimate, so $\int_{-2}^{1} x^2\,dx > L_3 = 5$ .”

What to write

“ $x^2$ decreases on $[-2, 0]$ and increases on $[0, 1]$ , so $L_3$ has no fixed side. In fact the integral is $3 < 5$ .”

x^2

[-2, 1]

with three left rectangles: the first one towers above the curve, so

L_3 = 5

overshoots the integral

3

Why: Left rectangles sit under the curve only where $f$ increases. On a function that turns, bound each monotonic piece separately, as the table above says.

6. Using an antiderivative when the question says use the definition

all the method marks: the right number alone earns at most 1 mark

What not to write

“By the definition, $\int_0^4 (16 - x^2)\,dx = \left[16x - \frac{x^3}{3}\right]_0^4 = \frac{128}{3}$ .”

What to write

“ $\Delta x = \frac{4}{n}$ , $x_i = \frac{4i}{n}$ , $R_n = 64 - \frac{32(n+1)(2n+1)}{3n^2} \to 64 - \frac{64}{3} = \frac{128}{3}$ .”

Why: The question tests the limit of the sum, not the value. The antiderivative may be used AFTERWARDS to check the answer, in a line marked as a check.

7. Taking the midpoint sum for the average of the left and right sums

the whole part

What not to write

“ $M_4 = \frac{L_4 + R_4}{2} = \frac{50 + 34}{2} = 42$ .”

What to write

“ $M_4 = f(\frac{1}{2}) + f(\frac{3}{2}) + f(\frac{5}{2}) + f(\frac{7}{2}) = \frac{63 + 55 + 39 + 15}{4} = 43$ .”

Why: The midpoint sum evaluates $f$ at NEW points, the centres of the strips. Averaging $L_n$ and $R_n$ is a different estimate, which is not the one asked for.

8. Concluding from equal left and right sums that the estimate is exact

the conclusion of the question

What not to write

“The power readings give $L_6 = R_6 = 28$ kWh, so the energy is exactly $28$ kWh.”

What to write

“ $L_6 - R_6 = \Delta t\left[P(6) - P(18)\right] = 0$ only because both end readings are $0$ . With $P$ rising then falling, the guaranteed bracket is $20.8 \le E \le 35.2$ kWh.”

Why: Two sums that share all their interior terms differ only by their end values. Their agreement says nothing about the error, which only a monotonicity argument can bound.

9. Ignoring the order of the limits

1 mark, and every later sign

What not to write

“ $\int_3^0 2x\,dx = 9$ , the area of the triangle under $y = 2x$ .”

What to write

“ $\int_3^0 2x\,dx = -\int_0^3 2x\,dx = -9$ .”

Why: From $3$ down to $0$ , $\Delta x = -\frac{3}{n}$ is negative, and every term of the sum changes sign. The area of the triangle is $9$ ; the integral written with reversed limits is $-9$ .

Which method to choose

Which tool, by the FORM of the question

Read the verb and the data of the question before writing anything: each form of question has its own tool

If estimate with n rectangles, or from a table of readings → write $\Delta x$ , list the $n$ sample points, add; then say over or under from the monotonicity

Example: $16 - x^2$ on $[0, 4]$ : $L_4 = 50$ , $R_4 = 34$ , $M_4 = 43$
If evaluate using the definition, or as a limit of Riemann sums → $R_n$ with $x_i = a + i\,\Delta x$ , powers of $i$ , the sum formulas, then $n \to \infty$

Example: $\int_1^3 (x^2 - 2x)\,dx = \lim R_n = \frac{2}{3}$
If evaluate by interpreting in terms of areas → draw the graph, cut into triangles, rectangles, trapezoids and pieces of discs, with signs

Example: $\int_0^3 \sqrt{9 - x^2}\,dx = \frac{9\pi}{4}$ , a quarter disc
If a limit of sums is given, find the integral → the factor like $\frac{c}{n}$ is $\Delta x$ , the expression $a + i\,\Delta x$ inside $f$ gives $a$ , and $b = a + n\,\Delta x$

Example: $\lim \sum_{i=1}^{n} \frac{2}{n}\left(1 + \frac{2i}{n}\right) = \int_1^3 x\,dx = 4$
If integrals of f over pieces are given → additivity, reversal of the limits, linearity

Example: $\int_0^8 (3g + 1) = 3(2\pi - 4) + 8 = 6\pi - 4$
If show that $A \le \int_a^b f \le B$ → the minimum and maximum of $f$ on $[a, b]$ , then $m(b - a) \le \int_a^b f \le M(b - a)$ ; split to sharpen

Example: $2 \le \int_0^2 \sqrt{1 + x^3}\,dx \le 6$

No branch of this chapter uses an antiderivative. When the question does not impose a method, geometry is the fastest route whenever the graph is made of lines and circles; otherwise, the limit of $R_n$ .

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Evaluating an integral using the definition

When to use it: The question says use the definition, use Riemann sums, or evaluate as a limit

1 Write $\Delta x = \frac{b - a}{n}$ and $x_i = a + i\,\Delta x$ , with the numbers of the question, $a$ included.
2 Compute $f(x_i)$ and write it as a combination of powers of $i$ , constants in front. When $a \neq 0$ , writing $f$ in powers of $x - a$ often makes the cross terms vanish.
3 Write $R_n = \sum_{i=1}^{n} f(x_i)\,\Delta x$ , pull every factor that does not depend on $i$ out of the sum, and apply the formulas, with $\sum_{i=1}^{n} c = cn$ .
4 Test the closed form at $n = 1$ : it must equal $(b - a)f(b)$ .
5 Let $n \to \infty$ , dividing each factor by one power of $n$ , and compare the sign of the result with the graph.

Concluding sentence

“By definition, $\int_1^3 (x^2 - 2x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n}\left(\frac{4i^2}{n^2} - 1\right)\frac{2}{n} = \lim_{n \to \infty}\left[\frac{4(n+1)(2n+1)}{3n^2} - 2\right] = \frac{2}{3}$ .”

The trap: Taking the limit inside the sum, term by term, before the formulas are applied: each term tends to $0$ and the sum of $n$ of them is not $0$ .

Marking: Typically 1 mark for the width and the sample point, 2 for the sum rewritten with the formulas, 1 for the limit, and the n = 1 test is what protects the other four.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The closed form at n = 1

With one rectangle, $R_1 = (b - a)f(b)$ . Put $n = 1$ in your formula for $R_n$ : the two must agree.

$\int_1^3 (x^2 - 2x)\,dx$ : $R_1 = 2f(3) = 6$ and $\frac{4 \cdot 2 \cdot 3}{3} - 2 = 6$ . A formula that gives $0$ has lost the $a$ of $x_i$ .
The box around the graph

The answer must lie between $m(b - a)$ and $M(b - a)$ , where $m$ and $M$ are the smallest and largest values of $f$ on the interval.

$\int_0^4 (16 - x^2)\,dx = \frac{128}{3} \approx 42.7$ lies between $0 \cdot 4 = 0$ and $16 \cdot 4 = 64$ . An answer of $\frac{64}{3}$ fails the next check.
The sign against the picture

Sketch the graph in ten seconds and compare the parts above and below the axis with the sign of your answer.

$x^2 - 4x$ is below the axis on $(2, 4)$ and above on $(4, 5)$ , the lower part being wider and deeper: $\int_2^5 (x^2 - 4x)\,dx = -3$ must be negative.
Right sum and left sum from the same formula

For a monotonic $f$ , $L_n - R_n = \Delta x\left[f(a) - f(b)\right]$ : the two closed forms must differ by exactly that.

$16 - x^2$ on $[0, 4]$ : $L_4 - R_4 = 50 - 34 = 16 = 1 \cdot (16 - 0)$ .

The typical problem, taken apart

Using the definition when a is not zero, with a negative answer

Use the definition of the integral, with right endpoints, to evaluate $\int_2^5 (x^2 - 4x)\,dx$ . Interpret the sign of the result.

No antiderivative and no calculator: every step must be justified as on a MATH 141 midterm.

y = x^2 - 4x

[2, 5]

: the region dips below the axis on

(2, 4)

and rises above it only on

(4, 5)

, so the integral must be negative.

Step 1

$\Delta x = \frac{5 - 2}{n} = \frac{3}{n}$ and $x_i = 2 + \frac{3i}{n}$ .

Why

The first line carries a method mark on its own, and the $2 +$ is the part most often dropped. Without it the sample points cover $[0, 3]$ .

Step 2

$x^2 - 4x = (x - 2)^2 - 4$ , so $f(x_i) = \frac{9i^2}{n^2} - 4$ .

Why

Rewriting $f$ around $a = 2$ makes $x_i - 2 = \frac{3i}{n}$ appear directly: the term in $i$ never shows up, which removes one sum and one chance of a sign error. Expanding $\left(2 + \frac{3i}{n}\right)^2 - 4\left(2 + \frac{3i}{n}\right)$ gives the same result more slowly.

Step 3

$R_n = \sum_{i=1}^{n}\left(\frac{9i^2}{n^2} - 4\right)\frac{3}{n} = \frac{27}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} - 12 = \frac{9(n+1)(2n+1)}{2n^2} - 12$ .

Why

The constant $-4 \cdot \frac{3}{n}$ is added $n$ times, giving $-12$ , which is $-4$ times the length $3$ of the interval. Writing $-\frac{12}{n}$ here is the costliest slip of the exercise.

Step 4

Test at $n = 1$ : $R_1 = 3f(5) = 3 \cdot 5 = 15$ , and $\frac{9 \cdot 2 \cdot 3}{2} - 12 = 27 - 12 = 15$ .

Why

Ten seconds, before the limit: the closed form is confirmed on a case computed independently, so any later error can only be in the limit itself.

Step 5

$\frac{(n+1)(2n+1)}{n^2} = \left(1 + \frac{1}{n}\right)\left(2 + \frac{1}{n}\right) \to 2$ , so $R_n \to 9 - 12 = -3$ .

Why

Dividing each factor by one power of $n$ makes the leading coefficient $2$ visible. The negative sign agrees with the figure: more of the region lies below the axis than above.

The conclusion, written out

“By definition, $\int_2^5 (x^2 - 4x)\,dx = \lim_{n \to \infty}\left[\frac{9(n+1)(2n+1)}{2n^2} - 12\right] = -3$ . The integral is negative because the part of the region below the axis, on $(2, 4)$ , outweighs the part above, on $(4, 5)$ .”

The classic mistake on this problem: Writing $x_i = \frac{3i}{n}$ , which computes $\int_0^3 (x^2 - 4x)\,dx = 9 - 18 = -9$ ; or writing $-\frac{12}{n}$ for the constant term, which gives $+9$ and a sign that the figure contradicts at once.

Learn by heart

• $\Delta x = \frac{b - a}{n}$ , $x_i = a + i\,\Delta x$ : the $a$ is never optional.
• $\sum_{i=1}^{n} c = cn$ , $\sum i = \frac{n(n+1)}{2}$ , $\sum i^2 = \frac{n(n+1)(2n+1)}{6}$ , $\sum i^3 = \left[\frac{n(n+1)}{2}\right]^2$ , from $i = 1$ only.
• $\int_a^b f(x)\,dx = \lim_{n \to \infty} R_n$ ; left, right and midpoint sums share the same limit.
• Increasing: $L_n$ under, $R_n$ over. Decreasing: the reverse. Turning: split first.
• An integral is a SIGNED area; an area is $\int |f|$ , split at the crossings.
• $\int_b^a f = -\int_a^b f$ , $\int_a^b c\,dx = c(b - a)$ , $m(b - a) \le \int_a^b f \le M(b - a)$ .
• Test every closed form at $n = 1$ : $R_1 = (b - a)f(b)$ .

Frequently asked questions

How do I evaluate a definite integral using the definition with Riemann sums?

Write the width of a strip, b minus a over n, and the right endpoint a plus i times that width. Put the endpoint into the function and expand in powers of i. Pull the constants out of the sum, close it with the formulas for the sums of 1, i, i squared and i cubed, and finally let n go to infinity. No antiderivative is used at any step.

Is the left Riemann sum an overestimate or an underestimate?

It depends on the function, not on the word left. If the function increases on the interval, the left sum is an underestimate and the right sum an overestimate. If it decreases, it is the reverse. If the function turns inside the interval, neither sum has a fixed side: split at the turning point and bound each piece separately.

Why can a definite integral be zero or negative when an area cannot?

Because a Riemann sum multiplies each width by the height of the function, with its sign. Rectangles below the horizontal axis are subtracted, so the integral is the area above the axis minus the area below. To find the area of a region, split the interval where the graph crosses the axis and add the pieces as positive numbers.

How do I recognize a limit of sums as a definite integral?

Look for the factor that behaves like a constant over n: it is the width of a strip, and the constant is b minus a. Inside the function, the expression a plus i times that width is the sample point, which gives the left end a. The right end is a plus n times the width. Several readings may work, and they all give the same value.

Is the midpoint sum the average of the left and right sums?

No. The midpoint sum evaluates the function at the centre of each strip, which are new points. The average of the left and right sums is a different estimate. On the function 16 minus x squared from 0 to 4 with four strips, the midpoint sum is 43 while the average of the left and right sums is 42.

Practise it

Corrected exercises: Riemann sums and the definite integral, MATH 141 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Next sheet The Fundamental Theorem of Calculus

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Revision sheet: Riemann sums and the definite integral (MATH 141)

The essentials

Three ingredients, written before any number

Signed area and the properties you may quote

The rules in table form

Over or under? It depends on the direction of f, never on left or right

The mistakes that cost marks

1. Dropping the left end from the sample point

2. Summing a constant once instead of n times

3. Keeping the wrong leading term of a sum of squares

4. Reporting an integral of zero as an area of zero

5. Believing the left sum always underestimates

6. Using an antiderivative when the question says use the definition

7. Taking the midpoint sum for the average of the left and right sums

8. Concluding from equal left and right sums that the estimate is exact

9. Ignoring the order of the limits

Which method to choose

Which tool, by the FORM of the question

How the answer is expected to be written

Evaluating an integral using the definition

Check before you hand in

The typical problem, taken apart

Using the definition when a is not zero, with a negative answer

Learn by heart

Frequently asked questions

How do I evaluate a definite integral using the definition with Riemann sums?

Is the left Riemann sum an overestimate or an underestimate?

Why can a definite integral be zero or negative when an area cannot?

How do I recognize a limit of sums as a definite integral?

Is the midpoint sum the average of the left and right sums?

Corrected exercises: Riemann sums and the definite integral, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?