The Definite Integral

Riemann sums only approximate an area. But the more rectangles you use, the better the approximation gets. If you take the limit as the number of rectangles goes to infinity, you get the exact value, and that limit has its own name and symbol: the definite integral. This page is about what the notation means and how to translate between a limit of sums and an integral.

Key ideas

Sigma notation

The Greek capital letter sigma, $\Sigma$, means “add up”. The expression

$$\sum_{k=1}^{n} a_k = a_1 + a_2 + \dots + a_n$$

adds the terms $a_k$ for $k = 1, 2, \dots, n$. For example, $\displaystyle\sum_{k=1}^{4} k^2 = 1 + 4 + 9 + 16 = 30$.

A right Riemann sum in sigma notation

Split $[a, b]$ into $n$ equal subintervals. Then

$$\Delta x = \frac{b - a}{n}, \qquad x_k = a + k\,\Delta x \quad (\text{the right endpoint of the } k\text{th subinterval}).$$

The right Riemann sum is

$$\sum_{k=1}^{n} f(x_k)\,\Delta x = \sum_{k=1}^{n} f\!\left(a + \frac{k(b - a)}{n}\right) \frac{b - a}{n}.$$

The definite integral

If $f$ is continuous on $[a, b]$, the limit of the Riemann sums as $n \to \infty$ exists, and it is the definite integral of $f$ from $a$ to $b$:

$$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k)\,\Delta x .$$

Left, right, and midpoint sums all have the same limit, so any of them can be used.

Symbol	Meaning
$\int$	a stretched S, for “sum”
$a$, $b$	the lower and upper limits of integration
$f(x)$	the integrand, the heights
$dx$	the “tiny width”; it tells you the variable is $x$

As $n$ grows, the right sums for $\int_0^2 x^2\,dx$ get closer to the exact value $\tfrac{8}{3} \approx 2.667$.

What the integral measures

The definite integral is a signed area: regions above the $x$-axis count as positive and regions below count as negative. If $f$ is a rate of change, the integral is the accumulated change, so its units are (units of $f$) × (units of $x$).

Reading a limit of a sum

To turn $\displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} f(x_k)\,\Delta x$ into an integral:

1. The factor that looks like $\dfrac{\text{something}}{n}$ outside $f$ is $\Delta x$. Its numerator is $b - a$.
2. Inside the function, $x_k = a + k\,\Delta x$. The constant part is $a$.
3. Then $b = a + (b - a)$, and the integrand is whatever is done to $x_k$.

The answer is not unique. For example, if $x_k = 1 + \tfrac{3k}{n}$, you can treat the function as “square it” on $[1, 4]$, or as “add $1$ and square it” on $[0, 3]$:

$$\lim_{n \to \infty} \sum_{k=1}^{n} \left(1 + \frac{3k}{n}\right)^2 \frac{3}{n} = \int_1^4 x^2\,dx = \int_0^3 (1 + x)^2\,dx .$$

Worked examples

Example 1: Writing a sum in sigma notation

Write the right Riemann sum for $\displaystyle\int_0^2 x^2\,dx$ with $n$ equal subintervals in sigma notation. Then evaluate it for $n = 4$.

Solution. $\Delta x = \dfrac{2 - 0}{n} = \dfrac{2}{n}$ and $x_k = 0 + \dfrac{2k}{n} = \dfrac{2k}{n}$:

$$R_n = \sum_{k=1}^{n} \left(\frac{2k}{n}\right)^2 \frac{2}{n}$$

For $n = 4$, the terms use $x_k = 0.5, 1, 1.5, 2$:

$$R_4 = 0.5\,(0.25 + 1 + 2.25 + 4) = 3.75$$

That’s the left graph in the figure above.

Example 2: From a limit to an integral

Write $\displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} \sqrt{1 + \frac{3k}{n}} \cdot \frac{3}{n}$ as a definite integral.

Solution. $\Delta x = \dfrac{3}{n}$, so $b - a = 3$. Inside the square root, $x_k = 1 + \dfrac{3k}{n}$, so $a = 1$ and $b = 4$. The function takes the square root:

$$\int_1^4 \sqrt{x}\,dx$$

Another correct answer is $\displaystyle\int_0^3 \sqrt{1 + x}\,dx$ (with $a = 0$ and the "$1 +$" moved into the function).

Example 3: Evaluating with geometry

Find $\displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} \sqrt{16 - \left(\frac{4k}{n}\right)^2} \cdot \frac{4}{n}$.

Solution. $\Delta x = \dfrac{4}{n}$ and $x_k = \dfrac{4k}{n}$, so $a = 0$ and $b = 4$:

$$\lim_{n \to \infty} \sum_{k=1}^{n} \sqrt{16 - \left(\frac{4k}{n}\right)^2} \cdot \frac{4}{n} = \int_0^4 \sqrt{16 - x^2}\,dx$$

The graph of $y = \sqrt{16 - x^2}$ is the top half of the circle $x^2 + y^2 = 16$ (radius $4$). From $x = 0$ to $x = 4$ it’s a quarter circle:

$$\int_0^4 \sqrt{16 - x^2}\,dx = \frac{1}{4}\pi(4)^2 = 4\pi$$

Example 4: From an integral to a limit

Write $\displaystyle\int_2^5 (x^3 - x)\,dx$ as the limit of a right Riemann sum.

Solution. $\Delta x = \dfrac{5 - 2}{n} = \dfrac{3}{n}$ and $x_k = 2 + \dfrac{3k}{n}$:

$$\int_2^5 (x^3 - x)\,dx = \lim_{n \to \infty} \sum_{k=1}^{n} \left[\left(2 + \frac{3k}{n}\right)^3 - \left(2 + \frac{3k}{n}\right)\right] \frac{3}{n}$$

Common mistakes

Taking $\Delta x$ as $b$. In Example 2, $\tfrac{3}{n}$ tells you the width of the interval is $3$, not that $b = 3$. Find $a$ from the constant in $x_k$, then $b = a + 3$.

Forgetting the $\Delta x$ factor. A Riemann sum is heights times widths. Without $\Delta x$ the limit usually doesn’t exist, and it isn’t a definite integral.

Mismatching the limits and the integrand. If you keep the "$1 +$" inside the function, the interval starts at $0$; if you take it out as $a = 1$, the function changes. Don’t do both, or you’ll get $\int_1^4 \sqrt{1 + x}\,dx$, which is wrong.

Treating the integral as total area. $\int_a^b f(x)\,dx$ is a signed area. Parts below the axis subtract.

Dropping the dx. The $dx$ is part of the notation. It shows which variable you’re integrating with respect to, and AP graders expect it.

Practice

1. (Warm-up) For $\displaystyle\int_{-1}^3 (x^2 + 1)\,dx$ with $n$ equal subintervals, write $\Delta x$ and the right endpoint $x_k$.

Solution

$$\Delta x = \frac{3 - (-1)}{n} = \frac{4}{n}, \qquad x_k = -1 + \frac{4k}{n}$$

2. (Warm-up) Evaluate $\displaystyle\sum_{k=1}^{5} (2k - 1)$.

Solution

$$1 + 3 + 5 + 7 + 9 = 25$$

3. (Warm-up) Write $\displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} \left(\frac{k}{n}\right)^3 \frac{1}{n}$ as a definite integral.

Solution

$\Delta x = \dfrac{1}{n}$ and $x_k = \dfrac{k}{n}$, so $a = 0$ and $b = 1$:

$$\int_0^1 x^3\,dx$$

4. (Core) Write $\displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} \ln\!\left(1 + \frac{2k}{n}\right) \frac{2}{n}$ as a definite integral.

Solution

$\Delta x = \dfrac{2}{n}$ and $x_k = 1 + \dfrac{2k}{n}$, so $a = 1$ and $b = 3$:

$$\int_1^3 \ln x\,dx$$

Also correct: $\displaystyle\int_0^2 \ln(1 + x)\,dx$.

5. (Core) Write $\displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} \left(-2 + \frac{3k}{n}\right)^2 \frac{3}{n}$ as a definite integral with the integrand $x^2$.

Solution

$\Delta x = \dfrac{3}{n}$ and $x_k = -2 + \dfrac{3k}{n}$, so $a = -2$ and $b = 1$:

$$\int_{-2}^{1} x^2\,dx$$

6. (Core) Write $\displaystyle\int_1^3 \frac{1}{x}\,dx$ as the limit of a right Riemann sum.

Solution

$\Delta x = \dfrac{2}{n}$ and $x_k = 1 + \dfrac{2k}{n}$:

$$\int_1^3 \frac{1}{x}\,dx = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{1 + \frac{2k}{n}} \cdot \frac{2}{n}$$

7. (Core) Evaluate $\displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} \left(1 + \frac{4k}{n}\right) \frac{4}{n}$ by writing it as an integral and using geometry.

Solution

$\Delta x = \dfrac{4}{n}$, $x_k = \dfrac{4k}{n}$ with the function $1 + x$, so the limit is $\displaystyle\int_0^4 (1 + x)\,dx$.

The region under $y = 1 + x$ from $0$ to $4$ is a trapezoid with parallel sides $1$ and $5$ and width $4$:

$$\int_0^4 (1 + x)\,dx = \frac{1 + 5}{2} \cdot 4 = 12$$

(Or read it as $\int_1^5 x\,dx$, a trapezoid with sides $1$ and $5$: also $12$.)

8. (Challenge) Use the formula $\displaystyle\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ to find a formula for the right sum $R_n$ for $\displaystyle\int_0^2 x^2\,dx$ from Example 1. Then take the limit as $n \to \infty$ to find the exact value of the integral.

Solution

$$\begin{aligned} R_n &= \sum_{k=1}^{n} \frac{4k^2}{n^2} \cdot \frac{2}{n} = \frac{8}{n^3} \sum_{k=1}^{n} k^2 \\ &= \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{4(n+1)(2n+1)}{3n^2} \end{aligned}$$

Check: $R_4 = \dfrac{4(5)(9)}{48} = 3.75$, which matches Example 1.

$$\int_0^2 x^2\,dx = \lim_{n \to \infty} \frac{4(2n^2 + 3n + 1)}{3n^2} = \frac{4 \cdot 2}{3} = \frac{8}{3}$$

9. (Challenge) Write $\displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} \sqrt{4 - \left(-2 + \frac{2k}{n}\right)^2} \cdot \frac{2}{n}$ as a definite integral in two different ways, and evaluate it.

Solution

$\Delta x = \dfrac{2}{n}$. With $x_k = -2 + \dfrac{2k}{n}$, $a = -2$ and $b = 0$:

$$\int_{-2}^{0} \sqrt{4 - x^2}\,dx$$

Or keep the $-2$ inside the function, with $x_k = \dfrac{2k}{n}$ on $[0, 2]$:

$$\int_0^2 \sqrt{4 - (x - 2)^2}\,dx$$

In the first form, $y = \sqrt{4 - x^2}$ is the top half of a circle of radius $2$, and $[-2, 0]$ is a quarter of it:

$$\frac{1}{4}\pi(2)^2 = \pi$$