The Central Idea
Suppose a quantity changes according to a function \(f(x)\) on an interval from \(a\) to \(b\).
Divide that interval into small pieces. On each piece:
- choose a representative input;
- evaluate the function there;
- multiply the function value by the width of the piece;
- add all of the resulting contributions.
The basic structure is
\[ \boxed{\text{total} \approx \sum (\text{local value})(\text{small width})}. \]This is the organizing idea behind Riemann sums.
Approximating a Continuous Total with a Finite Sum
A continuously varying quantity contains infinitely many local values. We cannot add those values one by one in the ordinary sense.
Instead, we first build a finite approximation.
If an interval is divided into \(n\) pieces, we estimate one contribution from each piece:
\[ f(x_1^*)\Delta x_1,\quad f(x_2^*)\Delta x_2,\quad \ldots,\quad f(x_n^*)\Delta x_n. \]Adding gives
\[ \boxed{ \sum_{i=1}^{n} f(x_i^*)\Delta x_i }. \]This is the general form of a Riemann sum.
The Interval of Accumulation
Riemann sums begin with a closed interval
\[ [a,b]. \]The number \(a\) is the starting input and \(b\) is the ending input.
The total interval length is
\[ \boxed{b-a}. \]The next step is to divide this interval into smaller subintervals.
Partitions
A partition divides the interval \([a,b]\) using points
\[ a=x_0The width of the \(i\)-th subinterval is
\[ \boxed{\Delta x_i=x_i-x_{i-1}}. \]Equal-Width Partitions
Introductory Riemann sums often use \(n\) subintervals of equal width.
Since the total interval length is \(b-a\), each width is
\[ \boxed{\Delta x=\frac{b-a}{n}}. \]The partition points are then
\[ x_i=a+i\Delta x \]for
\[ i=0,1,\ldots,n. \]Example: Finding the Subinterval Width
Divide
\[ [2,10] \]into \(4\) equal subintervals.
The width is
\[ \Delta x=\frac{10-2}{4}=2. \]The partition points are
\[ 2,\quad4,\quad6,\quad8,\quad10. \]Therefore, the four subintervals are
\[ [2,4],\quad[4,6],\quad[6,8],\quad[8,10]. \]Representative Points
On each subinterval
\[ [x_{i-1},x_i], \]choose a point
\[ x_i^* \]inside that subinterval.
The function value
\[ f(x_i^*) \]is used as a representative height for that piece.
The estimated contribution is then
\[ \boxed{f(x_i^*)\Delta x_i}. \]The Rectangle Interpretation
When \(f(x)\) is graphed, each term
\[ f(x_i^*)\Delta x_i \]can be visualized as the signed area of a rectangle.
Its width is
\[ \Delta x_i \]and its height is
\[ f(x_i^*). \]Therefore,
\[ \boxed{\text{rectangle contribution} =\text{height}\times\text{width}}. \]The General Riemann Sum
Adding the contribution from every subinterval gives
\[ \boxed{ \sum_{i=1}^{n} f(x_i^*)\Delta x_i }. \]This expression is a Riemann sum for \(f\) on \([a,b]\).
If all subintervals have equal width, then \(\Delta x_i=\Delta x\), so
\[ \boxed{ \sum_{i=1}^{n} f(x_i^*)\Delta x }. \]Reading the Sigma Notation
In
\[ \sum_{i=1}^{n} f(x_i^*)\Delta x, \]- \(\sum\) tells us to add a sequence of terms;
- \(i\) is the index;
- \(i=1\) is the first term;
- \(n\) is the final term;
- \(f(x_i^*)\) is the representative function value;
- \(\Delta x\) is the subinterval width.
Left Riemann Sums
A left Riemann sum uses the left endpoint of each subinterval as the representative point.
On the \(i\)-th subinterval,
\[ x_i^*=x_{i-1}. \]For equal widths,
\[ \boxed{ L_n=\sum_{i=1}^{n} f(x_{i-1})\Delta x }. \]Left Endpoints for an Equal Partition
Since
\[ x_i=a+i\Delta x, \]the left endpoint of the \(i\)-th subinterval is
\[ x_{i-1}=a+(i-1)\Delta x. \]Therefore,
\[ \boxed{ L_n= \sum_{i=1}^{n} f\!\left(a+(i-1)\Delta x\right)\Delta x }. \]Right Riemann Sums
A right Riemann sum uses the right endpoint of each subinterval.
Thus,
\[ x_i^*=x_i. \]For equal widths,
\[ \boxed{ R_n=\sum_{i=1}^{n} f(x_i)\Delta x }. \]Right Endpoints for an Equal Partition
The right endpoint of the \(i\)-th subinterval is
\[ x_i=a+i\Delta x. \]Therefore,
\[ \boxed{ R_n= \sum_{i=1}^{n} f\!\left(a+i\Delta x\right)\Delta x }. \]Midpoint Riemann Sums
A midpoint Riemann sum uses the midpoint of each subinterval.
For
\[ [x_{i-1},x_i], \]the midpoint is
\[ m_i=\frac{x_{i-1}+x_i}{2}. \]Thus,
\[ \boxed{ M_n=\sum_{i=1}^{n}f(m_i)\Delta x }. \]Midpoints for an Equal Partition
Since
\[ x_{i-1}=a+(i-1)\Delta x \]and
\[ x_i=a+i\Delta x, \]the midpoint is
\[ m_i=a+\left(i-\frac12\right)\Delta x. \]Therefore,
\[ \boxed{ M_n= \sum_{i=1}^{n} f\!\left(a+\left(i-\frac12\right)\Delta x\right)\Delta x }. \]Example: Left Sum for \(f(x)=x^2\)
Approximate the accumulation of
\[ f(x)=x^2 \]on
\[ [0,2] \]using \(4\) equal subintervals and left endpoints.
First compute the width:
\[ \Delta x=\frac{2-0}{4}=\frac12. \]The left endpoints are
\[ 0,\quad\frac12,\quad1,\quad\frac32. \]Calculating the Left Sum
The left sum is
\[ L_4= \left[ f(0)+ f\!\left(\frac12\right)+ f(1)+ f\!\left(\frac32\right) \right]\frac12. \]Since \(f(x)=x^2\),
\[ L_4= \left[ 0^2+ \left(\frac12\right)^2+ 1^2+ \left(\frac32\right)^2 \right]\frac12. \]Therefore,
\[ L_4= \left[ 0+\frac14+1+\frac94 \right]\frac12. \] \[ L_4= \frac72\cdot\frac12. \]Hence,
\[ \boxed{L_4=\frac74=1.75}. \]Example: Right Sum for the Same Function
Again use
\[ f(x)=x^2 \]on \([0,2]\) with \(n=4\).
The right endpoints are
\[ \frac12,\quad1,\quad\frac32,\quad2. \]Therefore,
\[ R_4= \left[ \left(\frac12\right)^2+ 1^2+ \left(\frac32\right)^2+ 2^2 \right]\frac12. \]So
\[ R_4= \left[ \frac14+1+\frac94+4 \right]\frac12 = \frac{15}{2}\cdot\frac12. \]Hence,
\[ \boxed{R_4=\frac{15}{4}=3.75}. \]Example: Midpoint Sum for the Same Function
The midpoints of the four subintervals are
\[ \frac14,\quad\frac34,\quad\frac54,\quad\frac74. \]Therefore,
\[ M_4= \left[ \left(\frac14\right)^2+ \left(\frac34\right)^2+ \left(\frac54\right)^2+ \left(\frac74\right)^2 \right]\frac12. \]The sum inside the brackets is
\[ \frac{1+9+25+49}{16} = \frac{84}{16} = \frac{21}{4}. \]Therefore,
\[ \boxed{M_4=\frac{21}{8}=2.625}. \]Comparing Left, Right, and Midpoint Estimates
For \(f(x)=x^2\) on \([0,2]\) with four equal subintervals, we found
\[ L_4=1.75, \] \[ M_4=2.625, \]and
\[ R_4=3.75. \]The exact definite integral, introduced formally in the next lesson, is
\[ \int_0^2 x^2\,dx=\frac83\approx2.667. \]In this example, the midpoint estimate is substantially closer than either endpoint estimate.
Increasing Functions: Left versus Right Sums
If \(f\) is increasing and positive on an interval, then each left-endpoint rectangle tends to use a height below the function later in its subinterval.
The left sum therefore tends to underestimate.
The right-endpoint rectangles use the larger endpoint values and tend to overestimate.
In this common situation,
\[ \boxed{L_n\leq \text{actual accumulation}\leq R_n}. \]Decreasing Functions: Left versus Right Sums
If \(f\) is decreasing and positive, the direction reverses.
Left endpoints use the larger values, so the left sum tends to overestimate.
Right endpoints use the smaller values, so the right sum tends to underestimate.
Thus,
\[ \boxed{R_n\leq \text{actual accumulation}\leq L_n} \]for a decreasing positive function under these conditions.
A Riemann Sum Is More Than a Rectangle-Area Trick
The rectangle picture is useful, but the deeper structure is
\[ \boxed{(\text{rate or density})(\text{small input change})}. \]For example:
- velocity times a short time interval approximates displacement;
- flow rate times a short time interval approximates volume;
- linear density times a short length approximates mass;
- marginal cost times a small production change approximates cost change.
Riemann sums are a general accumulation tool.
Units in a Riemann Sum
Units provide an important interpretation check.
Suppose \(v(t)\) is measured in meters per second and \(\Delta t\) is measured in seconds.
Then
\[ v(t_i^*)\Delta t \]has units
\[ \frac{\text{meters}}{\text{second}} \cdot \text{second} = \text{meters}. \]So the Riemann sum approximates displacement.
Example: Approximating Displacement from Velocity
Suppose a vehicle's velocity is sampled every second:
| \(t\) (s) | \(v(t)\) (m/s) |
|---|---|
| 0 | 4 |
| 1 | 6 |
| 2 | 7 |
| 3 | 9 |
| 4 | 10 |
Using left endpoints on \([0,4]\) with \(\Delta t=1\),
\[ L_4=(4+6+7+9)(1). \]Therefore,
\[ \boxed{L_4=26\text{ meters}}. \]Riemann Sums from Tabular Data
A formula for \(f\) is not required.
If values are measured experimentally or listed in a table, we can still approximate accumulation by multiplying sampled values by interval widths and adding.
This makes Riemann sums useful for real data, where an exact symbolic function may not be known.
Unequal Subinterval Widths
Equal widths are convenient but not required.
For a general partition,
\[ a=x_0The Riemann sum is then
\[ \boxed{ \sum_{i=1}^{n}f(x_i^*)\Delta x_i }. \]Each local contribution uses its own width.
Example: Unequal Widths
Suppose the partition is
\[ 0,\quad1,\quad3,\quad4. \]The subinterval widths are
\[ \Delta x_1=1,\qquad \Delta x_2=2,\qquad \Delta x_3=1. \]If representative values are
\[ 2,\quad5,\quad4, \]then the Riemann sum is
\[ 2(1)+5(2)+4(1). \]Therefore,
\[ \boxed{16}. \]Lower Sums
On each subinterval, choose the minimum value of \(f\).
If
\[ m_i=\min_{x\in[x_{i-1},x_i]}f(x), \]then the corresponding lower sum is
\[ \boxed{ \sum_{i=1}^{n}m_i\Delta x_i }. \]For a nonnegative function, this builds rectangles that do not rise above the graph on their respective subintervals.
Upper Sums
On each subinterval, choose the maximum value of \(f\).
If
\[ M_i=\max_{x\in[x_{i-1},x_i]}f(x), \]then the upper sum is
\[ \boxed{ \sum_{i=1}^{n}M_i\Delta x_i }. \]For a nonnegative function, these rectangles cover at least as much vertical extent as the graph on each subinterval.
Lower and Upper Bounds
For a bounded nonnegative function, lower and upper sums can bracket the accumulated area:
\[ \boxed{ \text{lower sum} \leq \text{area} \leq \text{upper sum} }. \]Refining the partition typically narrows the gap.
This bracketing idea is one route toward a rigorous definition of the integral.
Signed Contributions
If
\[ f(x_i^*)>0, \]then
\[ f(x_i^*)\Delta x_i>0. \]If
\[ f(x_i^*)<0, \]then the contribution is negative.
Thus a Riemann sum naturally represents signed accumulation, not necessarily ordinary geometric area.
Area Above and Below the Axis
Geometrically:
- rectangles above the \(x\)-axis contribute positively;
- rectangles below the \(x\)-axis contribute negatively.
A definite integral therefore measures net signed area.
If total geometric area is required, negative regions must later be handled using absolute values or by splitting the interval appropriately.
Why More Subintervals Usually Help
A wide rectangle represents the function by one height over a large interval.
If the function changes substantially across that interval, the rectangle may be a poor approximation.
By using more subintervals, each rectangle represents the function over a smaller region.
The local approximation generally improves as the widths shrink.
The Mesh of a Partition
For a partition with possibly unequal widths, the mesh is the largest subinterval width:
\[ \boxed{ \|P\|=\max_i \Delta x_i }. \]Saying that the partition becomes finer means that
\[ \|P\|\to0. \]This is the general version of making \(\Delta x\) smaller.
The Limiting Idea
A finite Riemann sum is an approximation.
The exact integral arises by considering what happens when the partition becomes arbitrarily fine.
Informally,
\[ \boxed{ \text{definite integral} = \lim_{\text{widths}\to0} \text{Riemann sum} }. \]For equal partitions, this is often written using \(n\to\infty\).
Preview of the Definite Integral
When the Riemann sums approach a unique limit independent of the particular representative points, we write
\[ \boxed{ \int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*)\Delta x }. \]This is the central bridge from finite sums to exact continuous accumulation.
Riemann Sum versus Definite Integral
These concepts are closely related but not identical.
A Riemann sum with finitely many subintervals is typically an approximation:
\[ \sum_{i=1}^{n}f(x_i^*)\Delta x. \]The definite integral is the limiting exact value:
\[ \int_a^b f(x)\,dx. \]Think:
\[ \boxed{\text{finite sum} \longrightarrow \text{limit} \longrightarrow \text{definite integral}}. \]Useful Equal-Partition Formulas
On \([a,b]\) with \(n\) equal subintervals:
\[ \boxed{\Delta x=\frac{b-a}{n}}. \]Partition points:
\[ \boxed{x_i=a+i\Delta x}. \]Left endpoints:
\[ \boxed{x_{i-1}=a+(i-1)\Delta x}. \]Midpoints:
\[ \boxed{m_i=a+\left(i-\frac12\right)\Delta x}. \]Example: Writing a Left Sum in Sigma Notation
Let
\[ f(x)=x^2+1 \]on
\[ [1,5] \]with \(n\) equal subintervals.
Then
\[ \Delta x=\frac{5-1}{n}=\frac4n. \]The left endpoint of the \(i\)-th interval is
\[ x_{i-1}=1+\frac{4(i-1)}{n}. \]Therefore,
\[ \boxed{ L_n= \sum_{i=1}^{n} \left[ \left(1+\frac{4(i-1)}{n}\right)^2+1 \right]\frac4n }. \]Example: Writing a Right Sum in Sigma Notation
For the same function and interval,
\[ f(x)=x^2+1,\qquad [1,5], \]the right endpoint is
\[ x_i=1+\frac{4i}{n}. \]Thus,
\[ \boxed{ R_n= \sum_{i=1}^{n} \left[ \left(1+\frac{4i}{n}\right)^2+1 \right]\frac4n }. \]Example: Writing a Midpoint Sum in Sigma Notation
The midpoint of the \(i\)-th subinterval is
\[ m_i= 1+\frac{4}{n}\left(i-\frac12\right). \]Therefore,
\[ \boxed{ M_n= \sum_{i=1}^{n} \left[ \left( 1+\frac{4}{n}\left(i-\frac12\right) \right)^2+1 \right]\frac4n }. \]Useful Finite-Sum Identities
Some Riemann-sum limits can be evaluated algebraically using formulas such as
\[ \sum_{i=1}^{n}1=n, \] \[ \sum_{i=1}^{n}i=\frac{n(n+1)}{2}, \] \[ \sum_{i=1}^{n}i^2= \frac{n(n+1)(2n+1)}{6}, \]and
\[ \sum_{i=1}^{n}i^3= \left[\frac{n(n+1)}{2}\right]^2. \]These formulas allow certain exact definite integrals to be derived directly from their Riemann sums.
Example: Building an Exact Limit for \(f(x)=x\)
Consider \(f(x)=x\) on \([0,1]\) using right endpoints.
With \(n\) equal subintervals,
\[ \Delta x=\frac1n. \]The right endpoints are
\[ x_i=\frac{i}{n}. \]So
\[ R_n= \sum_{i=1}^{n} \frac{i}{n}\frac1n. \]Hence,
\[ R_n= \frac{1}{n^2} \sum_{i=1}^{n}i. \]Taking the Limit
Use
\[ \sum_{i=1}^{n}i=\frac{n(n+1)}{2}. \]Then
\[ R_n= \frac{1}{n^2} \cdot \frac{n(n+1)}{2}. \]Simplify:
\[ R_n= \frac{n+1}{2n}. \]Taking \(n\to\infty\),
\[ \lim_{n\to\infty}R_n = \frac12. \]Therefore,
\[ \boxed{\int_0^1 x\,dx=\frac12}. \]Example: A Riemann-Sum Limit for \(x^2\)
On \([0,1]\), again use right endpoints.
Then
\[ \Delta x=\frac1n, \qquad x_i=\frac{i}{n}. \]Therefore,
\[ R_n= \sum_{i=1}^{n} \left(\frac{i}{n}\right)^2\frac1n. \]So
\[ R_n= \frac{1}{n^3} \sum_{i=1}^{n}i^2. \]Evaluating the \(x^2\) Limit
Use
\[ \sum_{i=1}^{n}i^2= \frac{n(n+1)(2n+1)}{6}. \]Then
\[ R_n= \frac{n(n+1)(2n+1)}{6n^3}. \]Divide numerator and denominator by \(n^3\):
\[ R_n= \frac{ (1)\left(1+\frac1n\right) \left(2+\frac1n\right) }{6}. \]Hence,
\[ \lim_{n\to\infty}R_n = \frac{1\cdot1\cdot2}{6} = \frac13. \]Therefore,
\[ \boxed{\int_0^1x^2\,dx=\frac13}. \]Recognizing a Riemann Sum from a Limit
Sometimes a problem gives a limit such as
\[ \lim_{n\to\infty} \sum_{i=1}^{n} \left(2+\frac{3i}{n}\right)^2 \frac3n. \]Compare this with
\[ \sum_{i=1}^{n}f(x_i)\Delta x. \]Here,
\[ \Delta x=\frac3n \]and
\[ x_i=2+\frac{3i}{n}. \]So the interval begins at \(2\) and ends at \(5\), and
\[ f(x)=x^2. \]Therefore the limit represents
\[ \boxed{\int_2^5x^2\,dx}. \]A Reliable Recognition Strategy
To convert a Riemann-sum limit into definite-integral notation:
- Identify the factor behaving like \(\Delta x\).
- Use \(\Delta x=(b-a)/n\) to determine the interval length.
- Identify the expression behaving like \(x_i\) or \(x_i^*\).
- Determine the starting point \(a\).
- Recover the function \(f\) from the expression being evaluated.
- Write \(\int_a^b f(x)\,dx\).
Indexing Matters
The expressions
\[ a+\frac{i(b-a)}{n} \]and
\[ a+\frac{(i-1)(b-a)}{n} \]are not interchangeable in a finite sum.
The first selects right endpoints. The second selects left endpoints.
Although both can converge to the same definite integral as \(n\to\infty\), they generally produce different finite approximations.
Using Monotonicity to Predict Error Direction
Before calculating, inspect whether the function is increasing or decreasing.
For a positive increasing function:
- left sums tend to be low;
- right sums tend to be high.
For a positive decreasing function:
- left sums tend to be high;
- right sums tend to be low.
This qualitative check can catch many numerical mistakes.
Why Midpoint Sums Often Perform Better
A left or right endpoint samples one edge of each subinterval.
The midpoint samples the center, which often balances variation within the interval more effectively.
This does not make midpoint sums exact in general, but for smooth functions they often provide much better accuracy than left or right sums using the same number of subintervals.
Averaging Left and Right Sums
For equal-width partitions, averaging the left and right sums gives the trapezoidal approximation:
\[ \boxed{ T_n=\frac{L_n+R_n}{2} }. \]The trapezoidal rule is developed more fully in the later Numerical Integration lesson.
It is useful to notice now that different sampling schemes can be combined to improve approximation.
Curvature Also Affects Approximation
Increasing versus decreasing behavior helps predict left- and right-sum error.
Concavity helps explain the behavior of midpoint and trapezoidal approximations.
Later numerical-integration methods quantify these errors more precisely. For now, the key idea is that the geometry of the function affects how well a chosen approximation represents it.
When a Riemann Sum Is Already Exact
If \(f(x)=c\) is constant on \([a,b]\), then every rectangle has exact height \(c\).
For equal widths,
\[ \sum_{i=1}^{n}c\Delta x = cn\Delta x. \]Since
\[ n\Delta x=b-a, \]we get
\[ \boxed{c(b-a)}. \]No limiting refinement is needed to improve the value.
Example: Accumulation from a Flow-Rate Table
Suppose water enters a tank at the following measured rates:
| \(t\) (min) | \(r(t)\) (L/min) |
|---|---|
| 0 | 8 |
| 2 | 10 |
| 4 | 9 |
| 6 | 12 |
Using a left sum over three subintervals of width \(2\) minutes,
\[ L_3=8(2)+10(2)+9(2). \]Therefore,
\[ \boxed{L_3=54\text{ liters}}. \]Why We Multiply by the Width
Adding only the sampled values
\[ f(x_1^*)+\cdots+f(x_n^*) \]does not account for how much input interval each value represents.
The factor
\[ \Delta x_i \]converts a local rate or density into a local accumulated amount.
That is why the building block is
\[ \boxed{f(x_i^*)\Delta x_i}, \]not merely \(f(x_i^*)\).
Common Error: Using \(\frac{b-a}{n-1}\)
If an interval is divided into \(n\) equal subintervals, the width is
\[ \boxed{\Delta x=\frac{b-a}{n}}. \]There are \(n+1\) partition points but only \(n\) subintervals.
Confusing the number of points with the number of intervals is a common source of error.
Common Error: Mixing Left and Right Endpoints
For a partition
\[ x_0,x_1,\ldots,x_n, \]a left sum uses
\[ x_0,x_1,\ldots,x_{n-1}, \]while a right sum uses
\[ x_1,x_2,\ldots,x_n. \]A quick check is that a left sum does not use \(b=x_n\), while a right sum does not use \(a=x_0\).
Common Error: Treating the Midpoint Index as an Endpoint
The midpoint of
\[ [x_{i-1},x_i] \]is
\[ \boxed{ \frac{x_{i-1}+x_i}{2} }. \]For equal partitions this becomes
\[ a+\left(i-\frac12\right)\Delta x, \]not \(a+i\Delta x\).
Common Error: Forgetting the Width Factor
The expression
\[ \sum_{i=1}^{n}f(x_i^*) \]is generally not the Riemann sum we need.
Each term must include the width:
\[ \boxed{ \sum_{i=1}^{n}f(x_i^*)\Delta x_i }. \]Common Error: Calling Every Riemann Sum “Area”
If \(f\) has units other than length, the product
\[ f(x_i^*)\Delta x_i \]may represent displacement, mass, volume, cost, probability, or another accumulated quantity.
Even geometrically, negative function values contribute signed area rather than ordinary positive area.
The broader term is accumulation.
Common Error: Treating a Finite Sum as the Integral
For finite \(n\),
\[ \sum_{i=1}^{n}f(x_i^*)\Delta x \]is generally an approximation.
The definite integral is obtained through a limiting process:
\[ \boxed{ \int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^{n}f(x_i^*)\Delta x }. \]A Reliable Procedure for a Numerical Riemann Sum
- Identify the interval \([a,b]\).
- Identify the number \(n\) of subintervals.
- Compute \(\Delta x=(b-a)/n\) if the widths are equal.
- List the partition points.
- Choose the required sample points: left, right, midpoint, or specified.
- Evaluate the function at those sample points.
- Multiply each value by its subinterval width.
- Add the contributions.
- Attach appropriate units.
- Interpret whether the result is an overestimate, underestimate, or neither when possible.
A Reliable Procedure for Building Sigma Notation
- Compute \(\Delta x=(b-a)/n\).
- Write the correct sample-point formula.
- Substitute that sample point into \(f\).
- Multiply by \(\Delta x\).
- Place the expression inside \(\sum_{i=1}^{n}\).
- Check that the first and last sample points match the intended rule.
Example: Full Right-Sum Workflow
Approximate the accumulation of
\[ f(x)=2x+1 \]on
\[ [1,5] \]using \(4\) right-endpoint rectangles.
First,
\[ \Delta x=\frac{5-1}{4}=1. \]The right endpoints are
\[ 2,\quad3,\quad4,\quad5. \]Completing the Right-Sum Calculation
Evaluate:
\[ f(2)=5,\qquad f(3)=7,\qquad f(4)=9,\qquad f(5)=11. \]Then
\[ R_4=(5+7+9+11)(1). \]Therefore,
\[ \boxed{R_4=32}. \]Because \(f(x)=2x+1\) is increasing, this right sum is an overestimate of the exact accumulation.
Example: Full Midpoint Workflow
Approximate
\[ f(x)=x^2+2 \]on \([0,4]\) with \(4\) midpoint rectangles.
The width is
\[ \Delta x=1. \]The midpoints are
\[ \frac12,\quad\frac32,\quad\frac52,\quad\frac72. \]Therefore,
\[ M_4= \left[ \left(\frac14+2\right) + \left(\frac94+2\right) + \left(\frac{25}{4}+2\right) + \left(\frac{49}{4}+2\right) \right](1). \]Simplifying gives
\[ \boxed{M_4=29}. \]Practice 1: Subinterval Width
Divide \([3,15]\) into \(6\) equal subintervals.
Find \(\Delta x\).
Practice 2: Partition Points
Divide \([0,6]\) into \(3\) equal subintervals.
List all partition points.
Practice 3: Left Sum
For
\[ f(x)=x+1 \]on \([0,4]\), use \(4\) equal subintervals and left endpoints to compute \(L_4\).
Practice 4: Right Sum
For the same function
\[ f(x)=x+1 \]on \([0,4]\), compute \(R_4\).
Practice 5: Midpoint Sum
For
\[ f(x)=x^2 \]on \([0,2]\), use \(2\) midpoint rectangles.
Practice 6: Overestimate or Underestimate?
Suppose \(f\) is positive and decreasing on \([a,b]\).
Which is generally larger: the left sum or the right sum?
Practice 7: Write a Right Sum
Write \(R_n\) in sigma notation for
\[ f(x)=x^3 \]on \([0,2]\) using \(n\) equal subintervals.
Practice 8: Write a Midpoint Sum
Write \(M_n\) in sigma notation for
\[ f(x)=2x+5 \]on \([1,4]\).
Practice 9: Recognize the Integral
Identify the definite integral represented by
\[ \lim_{n\to\infty} \sum_{i=1}^{n} \left(1+\frac{4i}{n}\right)^3 \frac4n. \]Practice 10: Tabular Accumulation
A rate \(r(t)\) is measured as follows:
| \(t\) | \(r(t)\) |
|---|---|
| 0 | 3 |
| 2 | 5 |
| 4 | 4 |
| 6 | 7 |
Use a left Riemann sum on \([0,6]\).
Practice 11: Unequal Widths
A partition has points
\[ 0,\quad2,\quad5,\quad6. \]Representative function values on the three subintervals are
\[ 4,\quad3,\quad8. \]Compute the corresponding Riemann sum.
Practice 12: Identify the Error
A student divides \([2,10]\) into \(4\) equal subintervals and writes
\[ \Delta x=\frac{10-2}{4-1}. \]Explain the error and give the correct width.
Practice Answers
Practice 1
\[ \Delta x=\frac{15-3}{6}=2. \]Therefore,
\[ \boxed{\Delta x=2}. \]Practice 2
The width is
\[ \Delta x=\frac{6-0}{3}=2. \]The partition points are
\[ \boxed{0,\ 2,\ 4,\ 6}. \]Practice 3
Here \(\Delta x=1\), and the left endpoints are
\[ 0,\quad1,\quad2,\quad3. \]Therefore,
\[ L_4= [f(0)+f(1)+f(2)+f(3)](1). \] \[ L_4=(1+2+3+4)(1). \]Hence,
\[ \boxed{L_4=10}. \]Practice 4
The right endpoints are
\[ 1,\quad2,\quad3,\quad4. \]Thus,
\[ R_4=(2+3+4+5)(1). \]Therefore,
\[ \boxed{R_4=14}. \]Practice 5
The width is
\[ \Delta x=1. \]The midpoints are
\[ \frac12,\quad\frac32. \]Therefore,
\[ M_2= \left[ \left(\frac12\right)^2+ \left(\frac32\right)^2 \right](1). \] \[ M_2=\frac14+\frac94. \]Hence,
\[ \boxed{M_2=\frac52}. \]Practice 6
For a positive decreasing function, left endpoints use the larger values.
Therefore,
\[ \boxed{L_n>R_n} \]unless the relevant values happen to be equal.
Practice 7
On \([0,2]\),
\[ \Delta x=\frac2n \]and the right endpoints are
\[ x_i=\frac{2i}{n}. \]Thus,
\[ \boxed{ R_n= \sum_{i=1}^{n} \left(\frac{2i}{n}\right)^3 \frac2n }. \]Practice 8
On \([1,4]\),
\[ \Delta x=\frac3n. \]The midpoint is
\[ m_i= 1+\left(i-\frac12\right)\frac3n. \]Therefore,
\[ \boxed{ M_n= \sum_{i=1}^{n} \left[ 2\left( 1+\left(i-\frac12\right)\frac3n \right)+5 \right]\frac3n }. \]Practice 9
We have
\[ \Delta x=\frac4n \]and
\[ x_i=1+\frac{4i}{n}. \]The interval is \([1,5]\), and the function is \(f(x)=x^3\).
Therefore,
\[ \boxed{\int_1^5x^3\,dx}. \]Practice 10
The three subintervals each have width \(2\).
A left sum uses the values at \(t=0,2,4\):
\[ L_3=3(2)+5(2)+4(2). \]Thus,
\[ \boxed{L_3=24}. \]Practice 11
The widths are
\[ 2,\quad3,\quad1. \]Therefore,
\[ 4(2)+3(3)+8(1)=8+9+8. \]Hence,
\[ \boxed{25}. \]Practice 12
Four subintervals means the total interval length is divided by \(4\), not by \(4-1\).
The correct width is
\[ \boxed{ \Delta x=\frac{10-2}{4}=2 }. \]What to Remember
A Riemann sum approximates continuous accumulation by adding local contributions:
\[ \boxed{ \sum_{i=1}^{n}f(x_i^*)\Delta x_i }. \]For \(n\) equal subintervals on \([a,b]\),
\[ \boxed{\Delta x=\frac{b-a}{n}}. \]The major sampling choices are:
- left endpoints;
- right endpoints;
- midpoints;
- upper values;
- lower values;
- arbitrary representative points.
The essential limiting idea is
\[ \boxed{ \int_a^b f(x)\,dx = \lim \sum f(x_i^*)\Delta x_i }. \]Riemann sums are therefore the bridge between ordinary finite addition and exact continuous accumulation.
Where We Go Next
Riemann sums show how a continuous total can be approximated by finite sums. The next lesson, Definite Integrals, turns that limiting process into a central mathematical object.
We will develop definite-integral notation, signed area, accumulation, orientation, interval properties, and the basic algebraic rules that make integrals practical to work with.