The definite integral as a limit

Infinitely many rectangles

Rectangles give estimates of an area, and more rectangles give better estimates. Could infinitely many rectangles give the area exactly? How would you add up infinitely many things?

For simple functions, the right sum with n rectangles can be written as a formula in n. Then you can take its limit as n\to\infty. Increase n below and watch the formula, the picture, and the sequence of sums all settle on one value.

Right sum with n rectangles

The right sum Rn as n grows (n on a log scale)

Integral
n
Right sum Rn
Limit as n → ∞
Gap
The right sum as a formula in n

The definition

Split [a,b] into n strips of width \Delta x=\frac{b-a}{n} and choose a sample point x_i^* in each. For a continuous function, the Riemann sums approach the same number however the sample points are chosen. That number is the definite integral:

\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i^*)\,\Delta x.

The integral sign is a stretched S, for sum. The function f(x) is the integrand, a and b are the limits of integration, and dx records the variable and the shrinking width. The result is a signed area: regions below the axis count as negative.

Sums you need

To evaluate the limit, you need closed forms for sums of powers:

\sum_{i=1}^{n}1=n,\qquad \sum_{i=1}^{n}i=\frac{n(n+1)}{2},\qquad \sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6},\qquad \sum_{i=1}^{n}i^3=\left(\frac{n(n+1)}{2}\right)^2.

A worked example

For \int_0^2x^2\,dx with right endpoints, \Delta x=\frac2n and x_i=\frac{2i}{n}, so

R_n=\sum_{i=1}^{n}\left(\frac{2i}{n}\right)^2\frac2n=\frac{8}{n^3}\sum_{i=1}^{n}i^2=\frac{8}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}=\frac43\left(1+\frac1n\right)\left(2+\frac1n\right).

As n\to\infty, both brackets approach their constant terms, so R_n\to\frac43\cdot1\cdot2=\frac83. The next lesson shows a much faster way to get the same answer, and explains why it works.

Solve it step by step

Evaluate each integral from the definition, using right endpoints. Choose the expressions, then give the limit. The score counts steps you answer correctly on the first try.

Problems solved0
First-try steps0 of 0

    Common mistakes

    Forgetting the starting point

    On [1,3], the right endpoints are x_i=1+\frac{2i}{n}, not \frac{2i}{n}. The sample points start at a.

    Taking the limit too early

    Simplify the sum into a formula in n first. A sum with n terms cannot have its limit taken term by term.

    Pulling i out of the sum

    Constants such as \frac{8}{n^3} can come outside the sum, because they do not depend on i. The index i cannot.