8.4 Integration as the limit of a sum — Learn

Prerequisites: sigma notation from Section 4.4; the area under a curve and the Fundamental Theorem of Calculus from Sections 8.1 and 8.2.

By the end of this section you can: write a definite integral as a sum of strips; compute the sum for a simple function using the standard sum formulae; and take the limit as the number of strips grows without bound.

1. Approximating an area with strips

Divide the interval from $a$ to $b$ into $n$ equal strips. Each strip has width

For the $i$th strip, take the height as the value of the function at the right end, $f(x_i)$. The area of that strip is about $f(x_i)\delta x$, and the sum of the strip areas approximates the area under the curve:

Each term is a thin rectangle. The rectangles do not fit the curve exactly, so $S_n$ is an estimate, not the exact area.

Example 1. Write a sum for $\displaystyle\int_0^3 (2x + 1)dx$ using $n$ equal strips.

Divide $[0, 3]$ into $n$ strips, so the width of each strip is

The right end of the $i$th strip is $x_i = i\delta x = \dfrac{3i}{n}$. Then

2. The limit as the number of strips grows

As $n$ increases, the strips become thinner and the stair-step top follows the curve more closely. In the limit $n \to \infty$ the approximation becomes exact:

This is what the integral sign means. The area is the limit of a sum of infinitely many infinitely thin strips. The value does not depend on whether the height is taken at the left end, the right end or the middle, because the strip width shrinks to zero.

Two standard sums are needed to compute a limit by hand:

3. Computing an integral as a limit

Example 2. Use the limit of a sum to show that $\displaystyle\int_0^1 x^2dx = \frac{1}{3}$.

Divide $[0, 1]$ into $n$ equal strips of width $\delta x = \dfrac{1}{n}$. The right end of strip $i$ is $x_i = \dfrac{i}{n}$, so the height is $x_i^2 = \dfrac{i^2}{n^2}$. Then

Simplify the fraction:

Now take the limit. Divide every term by $n^2$:

The limit is $\dfrac{1}{3}$, so $\displaystyle\int_0^1 x^2dx = \frac{1}{3}$. This matches the value found by the Fundamental Theorem of Calculus.

Example 3. Use the limit of a sum to show that $\displaystyle\int_0^a xdx = \frac{a^2}{2}$ for $a > 0$.

Divide $[0, a]$ into $n$ strips of width $\delta x = \dfrac{a}{n}$, with $x_i = \dfrac{ia}{n}$. Then

As $n \to \infty$ the fraction $\dfrac{n+1}{n}$ tends to $1$, so

Watch out. Three mistakes are common.

Forgetting that $\delta x$ depends on $n$. The width is $\dfrac{b-a}{n}$, so it changes with $n$. Treating it as a fixed number breaks the sum.

Taking the limit before simplifying. Simplify the sum into a single fraction in $n$ first, then let $n \to \infty$. Taking the limit early turns $\delta x$ into $0$ and every strip area into $0$.

Using the wrong sum formula. The sum of $i^2$ is $\dfrac{n(n+1)(2n+1)}{6}$, not $\left(\dfrac{n(n+1)}{2}\right)^2$. Re-check: after taking the limit, compare your answer with the value found by integration.

4. Try it: strips under a curve

Predict first. The figure shows rectangles under the curve $y = x^2$ from $x = 0$ to $x = 2$. Before you move the slider, predict whether the sum of the rectangle areas is larger or smaller than $\dfrac{8}{3}$, the exact area. Write down your prediction and one reason for it.

Test. Move the slider to increase the number of rectangles. Watch the sum in the readout as the rectangles get thinner.

Explain. Write two sentences. The first explains why the sum for this curve is an overestimate at every value of $n$. The second explains why the sum approaches the exact area as $n$ grows.

You should now be able to

• write a definite integral as a sum of strip areas;
• write the width $\delta x$ and the height $f(x_i)$ for a given interval;
• use the standard sums for $i$ and $i^2$;
• simplify the sum and take the limit as $n \to \infty$;
• explain why the limit gives the exact area.