Reading the Notation
A definite integral is written with several pieces of information grouped into one symbol. Reading each piece correctly tells you what quantity is being accumulated and over which interval.
In this notation, \(f(x)\) is the integrand, \(x\) is the variable of integration, \(a\) is the lower bound, and \(b\) is the upper bound. The bounds specify the interval \([a,b]\): accumulation is measured from the starting value \(a\) to the ending value \(b\).
The symbol \(\int\) signals an integral. The numbers or expressions written below and above it are the bounds. The expression following the integral sign and before \(dx\) is the integrand. The final \(dx\) names the variable of integration; it is part of the notation, not an instruction to multiply by a separate quantity called \(dx\).
For the introductory examples here, the lower bound is less than the upper bound, so \(a<b\). The order matters when identifying the bounds: the lower one is written below the integral sign, and the upper one is written above it. The interval contains every input value from the lower endpoint to the upper endpoint, including both endpoints.
What the Integrand and Variable Tell You
The integrand describes the quantity being accumulated as the variable moves across the interval. It might be a rate, a density, or a function whose graph is being considered. The variable of integration tells you which input changes as the accumulation is measured. For instance, in \(\int_1^4 (x^2+2)\,dx\), the changing input is \(x\), and the integrand is \(x^2+2\).
The variable of integration is sometimes called a dummy variable. Its name identifies the input inside this particular integral; another letter can serve the same role if it replaces that letter consistently. For example, \(\int_1^4 (x^2+2)\,dx\) and \(\int_1^4 (u^2+2)\,du\) describe the same integrand rule over the same numerical interval. This is a renaming of notation, not a change in the bounds or in what is being measured.
The variable name must be read together with the integrand and differential. In \(\int_1^4 (x^2+2)\,dx\), the expression depends on \(x\). If the same letter is replaced throughout, the expression and differential must agree: \(u^2+2\) goes with \(du\). A mismatched expression such as \(\int_1^4 (x^2+2)\,du\) does not follow the standard pattern being used here.
Why Bounds and Units Fit Together
The bounds tell you which input values are included in the accumulation. In a rate context, the integrand gives the rate at a particular input value, while the interval states the span over which that rate acts. A rate multiplied by a small change in the input has units of accumulated quantity. Adding such contributions across the interval therefore has the units of the quantity being accumulated.
For example, if \(v(t)\) is a velocity measured in meters per second and \(t\) is measured in seconds, then \(v(t)\,dt\) has units of meters. Accordingly, an integral of velocity with respect to time describes accumulated displacement and has units of meters. This unit check does not compute the integral; it checks that the notation matches the intended context.
Worked Example: Identify Every Part
Worked Example: Identify Every Part
Read the notation \(\int_2^5 (3x-1)\,dx\). Identify the integrand, variable of integration, bounds, and interval.
Step 1: Find the integrand. The expression after the integral sign and before \(dx\) is \(3x-1\). Therefore, the integrand is \(3x-1\).
Step 2: Find the variable. The differential is \(dx\), so the variable of integration is \(x\). The formula \(3x-1\) is expressed in terms of that same variable.
Step 3: Read the bounds. The number below the integral sign is \(2\), so the lower bound is \(2\). The number above the integral sign is \(5\), so the upper bound is \(5\).
Step 4: State the interval. Accumulation is measured over \([2,5]\), from input value \(2\) to input value \(5\). The interval width is \(5-2=3\).
Check: The lower endpoint \(2\) is less than the upper endpoint \(5\), and the listed interval starts at \(2\) and ends at \(5\). The integrand and differential both use \(x\), so the parts are consistent.
Worked Example: Interpret a Rate Integral
Worked Example: Interpret a Rate Integral
A tank's water level changes at rate \(r(t)=4t+2\) centimeters per minute, where \(t\) is time in minutes. Interpret \(\int_1^3 r(t)\,dt\), including its bounds and units.
Step 1: Identify the integrand and variable. The integrand is \(r(t)=4t+2\), the rate of change in water level. The differential \(dt\) shows that \(t\), measured in minutes, is the variable of integration.
Step 2: Identify the bounds and interval. The lower bound is \(1\) minute and the upper bound is \(3\) minutes. The interval is \([1,3]\), so the accumulation covers the time from minute \(1\) to minute \(3\).
Step 3: Check the units of the accumulated quantity. The rate has units of centimeters per minute, and the time input has units of minutes. Multiplying those units gives \((\text{centimeters per minute})(\text{minutes})=\text{centimeters}\). Thus the integral describes the change in water level over that time interval, measured in centimeters.
Step 4: Check the interpretation against the notation. The rate is expressed as a function of \(t\), the differential is \(dt\), and both bounds are time values. The interpretation therefore uses the correct variable, interval, and accumulated units. This is an interpretation, not a numerical evaluation of the integral.
Worked Example: Distinguish the Variable from a Parameter
Worked Example: Distinguish the Variable from a Parameter
In \(\int_0^2 (kx+1)\,dx\), treat \(k\) as a fixed constant. Identify the variable, integrand, bounds, and interval. Explain what changes if the notation is written using \(u\) instead.
Step 1: Separate the roles of the letters. The differential is \(dx\), so \(x\) is the variable of integration. The letter \(k\) is a fixed parameter in the integrand \(kx+1\); it is not the variable being ranged across the interval.
Step 2: Read the bounds. The lower bound is \(0\) and the upper bound is \(2\). The interval is \([0,2]\), with width \(2-0=2\).
Step 3: Rename the variable consistently. Replacing \(x\) by \(u\) throughout gives \(\int_0^2 (ku+1)\,du\). The integrand rule has the same form, the bounds remain \(0\) and \(2\), and \(k\) remains fixed.
Check: In the original notation, \(x\) appears in both the integrand and \(dx\); in the renamed notation, \(u\) appears in both the integrand and \(du\). The parameter \(k\) and the interval have not changed, so the renaming has not changed what the notation describes.
A Short Derivation: Why the Units Are Those of Accumulation
Suppose a rate \(r(t)\) measures change in a quantity \(Q\) per unit of time. Its units are \(Q\)-units per time-unit. Over a short time interval of length \(\Delta t\), the corresponding change is approximately the rate times the time length. The units of that product are
Now divide a larger time interval into several short pieces. Each rate-times-time contribution has \(Q\)-units, so adding the contributions still gives \(Q\)-units. Definite integral notation records the accumulation of the rate over the interval; its bounds specify the time endpoints, and \(dt\) identifies time as the input being traversed. Therefore, the integral's units must be \(Q\)-units. The unit conclusion follows from the units of the rate and the variable, rather than from evaluating the integral.
Common Mistakes in Reading Bounds
A frequent error is to treat the upper bound as the starting point simply because it is visually above the integral sign. For \(\int_2^5 f(x)\,dx\), a mistaken reading might say, “The interval runs from \(5\) to \(2\).” That reverses the locations shown in the notation: \(2\) is below the sign and is the lower bound; \(5\) is above it and is the upper bound. The interval specified by these bounds is \([2,5]\), not \([5,2]\).
Another error is to name the integrand as the entire expression including the differential. In \(\int_2^5 f(x)\,dx\), \(f(x)\) is the integrand, while \(dx\) identifies the variable. Keeping these roles distinct makes it easier to match the notation to the context and check units.
Check Your Understanding
For each question, identify the relevant parts of the notation before giving an interpretation.
- In \(\int_1^4 (x+3)\,dx\), what are the integrand, variable of integration, lower bound, upper bound, and interval?
- A rate \(p(s)\) is measured in liters per hour. What units does \(\int_2^6 p(s)\,ds\) have, and which interval of the input is covered?
- In \(\int_0^3 (cu^2+1)\,du\), if \(c\) is fixed, which letter is the variable of integration and which is a parameter?
- Rewrite \(\int_2^5 (x^2-1)\,dx\) using \(u\) as the variable of integration. Which parts of the notation must change, and which stay the same?
- In \(\int_3^7 g(t)\,dt\), a student says the lower bound is \(7\). Identify the error and state the interval correctly.
