Area as the limit of strips

What you will be able to do

• cut the region under a curve into thin strips and add their areas;
• explain why thinner strips give a better answer, and why the limit is the definite integral;
• find the area under a curve with an antiderivative, and check it against a strip sum.

A lawn with a curved edge

A gardener wants to buy turf for a lawn. One side is a straight path. The other side is a curve. A rectangle is easy, because the area is length times width. A curved edge is harder. So the gardener marks the lawn into narrow strips. Each strip is almost a rectangle, and the areas of the strips add up to almost the area of the lawn.

This chapter uses the rules of integration from the previous chapter. We start by seeing why an integral gives an area at all.

Strips under a curve

Let $y = f(x)$ be continuous on $[a, b]$ and never below the $x$-axis. Cut $[a, b]$ into $n$ equal pieces. Each piece has the width

The symbol $\Delta x$ is read "delta x" and means a small step along the $x$-axis. On each piece, draw a rectangle. Its height is the height of the curve at the left end of the piece. The first rectangle has height $f(a)$, the next has height $f(a + \Delta x)$, and so on. The sum of the areas of all the rectangles is

You may also take the height at the right end or at the middle of each piece. Every such choice gives a sum that is close to the area when $n$ is large.

From the sum to the integral

With three strips the steps are far from the curve. With twelve strips they hug the curve. As $n$ grows without bound, the sum settles on a number. That number is the area, and it is the definite integral:

AP and IB texts call a sum of this kind a Riemann sum. In the Indian syllabus the same limit is the definition of the definite integral, met in the previous chapter.

Finding the area without a limit

Taking a limit of sums is slow. The Fundamental Theorem of Calculus gives a shortcut. If $F$ is an antiderivative of $f$, so that $F'(x) = f(x)$, then

The constant of integration cancels in the subtraction, so we leave it out. The area is a number with the unit of length times length, which is a square unit.

A worked example with the strips

Find the area under $y = x^2 + 1$ from $x = 0$ to $x = 3$, first with strips and then with the integral.

1. Three strips. Here $\Delta x = 1$. The left-end heights are $f(0) = 1$, $f(1) = 2$ and $f(2) = 5$. The sum is $1 + 2 + 5 = 8$.
2. More strips. With $6$ strips the sum is $9.875$. With $12$ strips it is $10.906$. The sums rise towards a limit.
3. The integral. An antiderivative of $x^2 + 1$ is $\dfrac{x^3}{3} + x$. So the area is $\left(\dfrac{27}{3} + 3\right) - 0 = 12$.

The strip sums are all below $12$ because the curve rises, so every left-end rectangle lies under the curve. The gap is $4$ for three strips and about $2.1$ for six. Doubling the strips roughly halves the gap.

A second example, and a check by shape

Find the area under $y = 3x^2$ between $x = 1$ and $x = 4$. An antiderivative is $x^3$, so the area is $4^3 - 1^3 = 63$ square units.

Now test the method on a shape we know. The line $y = 2x$ from $x = 0$ to $x = 4$ together with the $x$-axis and the line $x = 4$ makes a triangle. Its base is $4$ and its height is $8$, so its area is $\tfrac{1}{2} \times 4 \times 8 = 16$. The integral gives $\left[x^2\right]_0^4 = 16$. The two methods agree.

Practise with new numbers

Try it in Python

The program finds the left sum and the right sum for the curve $y = x^2 + 1$ with more and more strips. At the end it prints the exact area from the antiderivative.

Common mistakes

• Using the number of strips as the width. The width is $\dfrac{b - a}{n}$, not $n$.
• Counting one strip too many. A left-end sum uses the heights at $x_0, \dots, x_{n-1}$. The last end point is not used.
• Putting the constant back. In a definite integral the constant of integration cancels. Do not add $+ C$.
• Giving the answer without a unit. An area is in square units, such as square metres.

Remember these

• The area under $y = f(x)$ from $a$ to $b$ is the limit of the strip sums, and it is $\displaystyle\int_a^b f(x)\,dx$ when $f \ge 0$.
• Each strip has width $\Delta x = \dfrac{b - a}{n}$ and height $f(x_i)$.
• With an antiderivative $F$, the area is $F(b) - F(a)$.
• More strips give a closer answer, but the integral gives the exact one.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The interval $[1, 5]$ is cut into $8$ equal strips. What is the width of each strip?

2. For $y = 2x + 1$ on $[0, 3]$, find the sum of three left-end strips and the exact area. Is the sum above or below the exact area?

3. Find $\displaystyle\int_0^2 (3x^2 + 1)\,dx$.

4. Why does a larger number of strips give a better estimate of the area?

5. Find the area under $y = \sqrt{x}$ from $x = 0$ to $x = 9$.

References

• OpenStax, Calculus Volume 1, Chapter 5, sections on approximating areas and the definite integral.
• NCERT, Mathematics Part II for Class XII, Chapter 7, section on the definite integral as the limit of a sum.