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Calculus AB · Unit 6 · 6.3
AP Calculus · Integration and Accumulation of Change
Riemann Sums and Definite Integral Notation
Every symbol in the integral replaces one piece of the sum. Read them side by side and the notation explains itself.
key idea$\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^{n} f(x_i^*)\,\Delta x$
What it says
The definite integral as a limit
\[\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^{n} f\!\left(x_i^*\right)\Delta x\]
Slice $[a,b]$ into $n$ pieces of width $\Delta x=\dfrac{b-a}{n}$, build a rectangle on each, and add. As $n\to\infty$ the rectangles shrink and the sum converges to a single number — the definite integral. Every piece of the notation traces back to that picture.
Why the notation looks the way it does
The $\int$ is an elongated S for sum. The $dx$ is what $\Delta x$ becomes in the limit. The $f(x)$ between them is the rectangle's height. The integral sign is a picture of the process.
| Riemann sum piece | Integral piece | Means |
|---|
| $\sum$ | $\displaystyle\int$ | add up the pieces |
| $i=1$ to $n$ | from $a$ to $b$ | the interval covered |
| $f\!\left(x_i^*\right)$ | $f(x)$ | the height of a slice |
| $\Delta x$ | $dx$ | the width of a slice |
| $\lim_{n\to\infty}$ | — built in | the slices become infinitely thin |
Read the two columns side by side and the integral notation stops being arbitrary. Each symbol replaces one part of the sum.
Reading the notation
$\Delta x=\dfrac{b-a}{n}$The width of each slice when the partition is regular (equal widths).
$x_i^*$The sample point in slice $i$ where the height is measured. Left, right or midpoint.
$\displaystyle\sum_{i=1}^{n}$Summation notation: add the expression for $i=1,2,\dots,n$.
$\displaystyle\int_a^b f(x)\,dx$The definite integral: a number, the signed area between the curve and the axis on $[a,b]$.
$dx$Marks the variable of integration. It also tells you which letter is the input — essential when other letters appear.
Dummy variable$\int_a^b f(x)\,dx$ and $\int_a^b f(t)\,dt$ are the same number. The letter is internal.
$\int$an elongated S for sum
$dx$what $\Delta x$ becomes as the slices vanish
the limitswhich interval is being sliced
Worked examples
EX 1
Sum to integral
recognise the limitExpress $\lim\limits_{n\to\infty}\sum\limits_{i=1}^{n}\left(1+\dfrac{3i}{n}\right)^{2}\cdot\dfrac{3}{n}$ as a definite integral.
1
Find $\Delta x$
The trailing factor is $\dfrac{3}{n}$, so $\Delta x=\dfrac{3}{n}$ and therefore $b-a=3$.
2
Find the sample point
$x_i^*=1+\dfrac{3i}{n}$. At $i=0$ this is $1$; at $i=n$ it is $4$.
3
Read off the limits
$a=1$ and $b=4$, consistent with $b-a=3$. ✓
4
Read off the integrand
The height is $\left(x_i^*\right)^{2}$, so $f(x)=x^{2}$.
5
Write it
$\displaystyle\int_{1}^{4}x^{2}\,dx$.
EX 2
Integral to sum
the other directionWrite $\displaystyle\int_{0}^{2}e^{x}\,dx$ as the limit of a right Riemann sum.
1
Compute $\Delta x$
$\dfrac{b-a}{n}=\dfrac{2-0}{n}=\dfrac{2}{n}$.
2
Write the right sample point
$x_i=a+i\Delta x=\dfrac{2i}{n}$ for $i=1,\dots,n$.
3
Substitute into $f$
$f\!\left(x_i\right)=e^{2i/n}$.
4
Assemble
$\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}e^{2i/n}\cdot\dfrac{2}{n}$.
5
Check the index range
A right sum runs $i=1$ to $n$; a left sum would run $i=0$ to $n-1$. That single change is the difference between them.
EX 3
The dummy variable
notationAre $\displaystyle\int_{0}^{3}t^{2}\,dt$, $\displaystyle\int_{0}^{3}u^{2}\,du$ and $\displaystyle\int_{0}^{3}x^{2}\,dt$ the same?
1
Compare the first two
Both integrate the squaring function from $0$ to $3$. The letter is internal, so they are identical numbers.
2
Examine the third
The integrand uses $x$ but $dt$ says the variable of integration is $t$.
3
Treat $x$ as a constant
With respect to $t$, $x^{2}$ is a constant, so the integral is $x^{2}\cdot(3-0)=3x^{2}$.
4
State the difference
The first two are the number $9$; the third is the function $3x^{2}$.
5
The lesson
$dx$ is not decoration. It names the variable, and mismatching it changes the answer's type.
EX 4
Evaluate by geometry
no antiderivative neededEvaluate $\displaystyle\int_{-3}^{3}\sqrt{9-x^{2}}\,dx$ using the definition.
1
Identify the curve
$y=\sqrt{9-x^{2}}$ is the upper half of the circle $x^{2}+y^{2}=9$.
2
Recall what the integral means
The signed area between the curve and the axis from $-3$ to $3$.
3
Compute the area geometrically
A semicircle of radius $3$ has area $\tfrac12\pi r^{2}=\tfrac{9\pi}{2}$.
4
State the answer
$\displaystyle\int_{-3}^{3}\sqrt{9-x^{2}}\,dx=\dfrac{9\pi}{2}$.
5
Why this works
The integral is the area. When the region is a familiar shape, geometry beats any antiderivative — and here no elementary antiderivative would help.
Boundary cases
A definite integral is a number
$\int_a^b f(x)\,dx$ evaluates to a single value. It is not a function of $x$ — $x$ is internal to the notation.
The $dx$ names the variable
$\int_0^3 x^{2}\,dt$ treats $x$ as constant and returns $3x^{2}$. Mismatching the letter changes the answer's type entirely.
Left and right sums differ only in the index range
$i=1$ to $n$ is a right sum; $i=0$ to $n-1$ is a left sum. Same expression, shifted window.
Signed, not absolute
Regions below the axis contribute negatively. The integral is signed area, so it can be negative or zero.