IB Analysis and Approaches HL · Integral calculus · Section 9 of 13

Definite integral and signed area under a curve

3 worked examples · By Álvaro López

Key idea

The definite integral $\int_{a}^{b} f (x) d x$ measures the "signed area" between the graph of $f$ and the $x$ -axis, from $x = a$ to $x = b$ . It is defined as the limit of sums of increasingly thin rectangles (Riemann sums). It is the basis of areas, volumes and the applications in the rest of the topic.

Approximating an area with rectangles

To find the area under a curve there is no elementary geometric formula. The idea is to cover it with rectangles, whose areas you can calculate.

Divide $[a, b]$ into $n$ subintervals of equal width $Δ x = \frac{b - a}{n}$ , with endpoints $x_{0} = a, x_{1}, \dots, x_{n} = b$ . In each subinterval $[x_{i - 1}, x_{i}]$ you can draw two rectangles:

one inscribed, with height equal to the minimum value of $f$ on that subinterval, $m_{i}$ ;
one circumscribed, with height equal to the maximum value, $M_{i}$ .

The sum of the areas of the first set is the lower sum and that of the second set is the upper sum: $L_{n} = i = 1 \sum n m_{i} Δ x, U_{n} = i = 1 \sum n M_{i} Δ x .$

As the inscribed rectangles lie inside the region and the circumscribed ones surround it, the area $A$ satisfies $L_{n} \leq A \leq U_{n}$ .

Graph of y=x^2 on (0,2) with four inscribed rectangles (filled) and four circumscribed rectangles (dashed outline)

If $f$ is increasing, the minimum is at the left endpoint and the maximum at the right endpoint, and the sums are simple to write. If it is decreasing, the opposite happens.

From sums to the integral

As $n$ increases, the lower sum grows and the upper sum decreases, and both get closer to the same number. This is what is called the definite integral.

A complete example with $f (x) = x^{2}$ on $[0, 1]$ ( $Δ x = \frac{1}{n}$ ; increasing function, so the minimum is at the left endpoint): $L_{n} = i = 1 \sum n (\frac{i - 1}{n})^{2} \frac{1}{n} = \frac{( n - 1 ) n ( 2 n - 1 )}{6 n ^{3}} n \to \infty \frac{1}{3} .$ This confirms that $\int_{0}^{1} x^{2} d x = \frac{1}{3}$ , the same value you will obtain with the fundamental theorem of calculus in the next section.

Signed area

If $f$ takes negative values, the rectangles below the axis have a "negative height", and the integral counts them with a minus sign. That is why the definite integral is a signed area: $\int_{a}^{b} f (x) d x = (area above the axis) - (area below the axis) .$ This difference can be zero, or even negative, even though the region is not empty. For the actual area more care is needed: you will see this in the section on calculating areas.

When the region is a trapezium, a triangle, a rectangle or part of a circle, you can find the integral without calculating antiderivatives, using only geometry formulae. This is a common trick on Paper 1.

Worked examples

Example 1 Basic

Lower and upper sums

Let $f (x) = x^{2}$ on the interval $[0, 2]$ and take $n = 4$ subintervals of equal width.

a) Calculate the lower sum $L_{4}$ .

b) Calculate the upper sum $U_{4}$ .

c) Give two numbers between which $\int_{0}^{2} x^{2} d x$ lies.

Solution

The width is $Δ x = \frac{2 - 0}{4} = 0.5$ , and the endpoints are $0; 0.5; 1; 1.5; 2$ . As $x^{2}$ is increasing on $[0, 2]$ , the minimum is at the left endpoint and the maximum at the right endpoint.

a) $L_{4} = 0.5 (0^{2} + 0. 5^{2} + 1^{2} + 1. 5^{2}) = 0.5 (0 + 0.25 + 1 + 2.25) = \frac{7}{4}$ .

b) $U_{4} = 0.5 (0. 5^{2} + 1^{2} + 1. 5^{2} + 2^{2}) = 0.5 (0.25 + 1 + 2.25 + 4) = \frac{15}{4}$ .

c) $\frac{7}{4} \leq \int_{0}^{2} x^{2} d x \leq \frac{15}{4}$ , that is, between $1.75$ and $3.75$ . (The exact value is $\frac{8}{3} \approx 2.67$ .)

Answer: a) $L_{4} = \frac{7}{4}$ ; b) $U_{4} = \frac{15}{4}$ ; c) the integral lies between $\frac{7}{4}$ and $\frac{15}{4}$ .

Example 2 Medium

Integrals with geometry

Calculate, without finding antiderivatives, by drawing a sketch of the region:

a) $\int_{0}^{4} 16 - x^{2} d x$ .

b) $\int_{- 3}^{3} (2 - ∣ x ∣) d x$ .

c) $\int_{0}^{4} (x - 2) d x$ .

Solution

a) The graph of $y = 16 - x^{2}$ is the upper semicircle of radius $4$ centred at the origin. Between $x = 0$ and $x = 4$ it is a quarter circle, with area $\frac{1}{4} π \cdot 4^{2} = 4 π$ .

b) The graph of $y = 2 - ∣ x ∣$ is a triangle with vertex $(0, 2)$ that cuts the $x$ -axis at $\pm 2$ . Between $- 2$ and $2$ it lies above the axis, with area $\frac{1}{2} \cdot 4 \cdot 2 = 4$ . Between $2$ and $3$ it lies below the axis: it is a triangle with base $1$ and height $1$ , with area $\frac{1}{2}$ , and the same between $- 3$ and $- 2$ . Therefore the integral is $4 - \frac{1}{2} - \frac{1}{2} = 3$ .

c) The line $y = x - 2$ passes through $(2, 0)$ . Between $0$ and $2$ it forms a triangle below the axis of area $\frac{1}{2} \cdot 2 \cdot 2 = 2$ , and between $2$ and $4$ an identical one above the axis. The integral is $2 - 2 = 0$ .

Answer: a) $4 π$ ; b) $3$ ; c) $0$ .

Example 3 Hard

Distance from a table of velocities

A car accelerates for $40$ seconds. Its velocity $v$ (in m/s), measured every $10$ seconds, is:

$t$ (s)	0	10	20	30	40
$v$ (m/s)	0	8	14	18	20

The velocity is increasing over that interval. The distance travelled is $\int_{0}^{40} v (t) d t$ .

a) Find a lower bound for the distance using rectangles with the height at the left endpoint.

b) Find an upper bound using the height at the right endpoint.

c) Estimate the distance as the mean of the two bounds.

d) What is the difference between the two bounds? How could you reduce it?

Solution

The width of each rectangle is $Δ t = 10$ s. As $v$ is increasing, the left endpoint gives the minimum on each interval and the right endpoint the maximum.

a) $L = 10 (0 + 8 + 14 + 18) = 400$ m.

b) $U = 10 (8 + 14 + 18 + 20) = 600$ m.

c) The mean is $\frac{400 + 600}{2} = 500$ m (this is equivalent to approximating the region with trapezia).

d) $U - L = 200$ m. As $v$ is increasing, the difference is $Δ t (v (40) - v (0)) = 10 \cdot 20 = 200$ . To reduce it you must measure more often: with $Δ t = 5$ s the difference falls to $100$ m.

Answer: a) $400$ m; b) $600$ m; c) $500$ m; d) $200$ m.

In the exam: in this type of question justify why each endpoint gives a bound (the function is increasing), and write out the full sum, with the width of each rectangle, before giving the result.

Common mistakes

Summary

Lower $L_{n}$ and upper $U_{n}$ sums: rectangles of width $Δ x = \frac{b - a}{n}$ with minimum or maximum height in each subinterval; $L_{n} \leq A \leq U_{n}$ .
$\int_{a}^{b} f (x) d x = n \to \infty lim L_{n} = n \to \infty lim U_{n}$ if $f$ is continuous on $[a, b]$ .
It is a signed area: what lies below the axis counts negatively.
If $f \geq 0$ , the integral is the area under the curve; if $f \leq 0$ , it is the same area with a negative sign.
With known geometric regions you can calculate the integral without antiderivatives.

Keep studying

Maths classes in Aldaia and online

Want to work on it together?

Free first consultation: we look at your level and your exam, and build a plan. Álvaro replies on WhatsApp.

Message us on WhatsApp