# Area under a curve inquiry

# The prompt

**Mathematical inquiry processes:**** **Explore; generate examples; test cases; generalise and prove.** ****Conceptual field of inquiry: **Trapezium rule; integration; definite integrals; points of intersection.

The prompt has been used in classroom inquiries to introduce the concept of the definite integral as the area under a curve between two limits.

The prompt consists of two parts - a statement and a diagram. The statement asserts that the ratio of the area under the straight line to the area under the curve is 2:1. The diagram suggests that the areas are bounded by the* x*-axis and the vertical line(s) produced from the point(s) of intersection to the *x*-axis. The teacher should confirm that the *x*-axis will always be the boundary throughout the inquiry. The vertical lines, however, might be linked to the points of intersection only in the initial line of inquiry.

Students do not need any knowledge of integration to ask questions and make observations about the statement and diagram. These are some that have arisen in classroom inquiry:

Can the statement be true?

Are the areas measured above the

*x*-axis?The straight line and the curve intersect at (0,0)?

What is the other point of intersection?

To find the points of intersection, you make the functions equal.

The curve could be

*y*=*x*^{3}.The straight line looks like

*y*= 3*x*.The area under the straight line is a triangle.

How do you work out the area of the region under the curve?

You can estimate the area under the curve using the trapezium rule.

How can you prove the statement is true?

The equations of the straight line and curve in the prompt are *y *= 3*x* and y = *x*^{3} - 3*x*^{2} + 3*x *respectively. By integrating both separately, the teacher can demonstrate that the ratio of the areas is 2:1 as required in the prompt.

As the inquiry develops, the teacher must judge the best moment to explain why the definite integral gives the area under the curve. That might be after the initial phase of questioning and noticing or it could be delayed until students have practised integrating.

# Student leaders of inquiry

When a class carries out an inquiry, a new culture in which students take on leadership roles can quickly develop. The initiative shown by students who have been quiet in previous lessons can come as a surprise to the teacher.

Moreover, the new leaders have an impact on the rest of the class. Their deep engagement with the mathematics, their resilience in pursuit of a line of inquiry and their willingness to explain an approach inspire their peers.

This was the case with two year 12 girls, Roisin and Cath. After discussing the prompt, the students in the class decided to start the inquiry by creating examples with y = *x*^{3}. They changed the coefficient of *x* in the linear equation.

Roisin noticed that the results were always the same - the ratio of the area under the line to the area under the curve is 2:1 whatever the gradient of the line. She set out independently to prove the generalisation (see her working below). At the end of the lesson, Roisin presented her line of inquiry to the class.

Roisin's proof of her generalisation.

Cath's proof that the ratio of the area under the line *y *= *mx *to the area under the curve* **y* =* x*^{n}* *is (*n* + 1):2

Roisin's approach generated a lot of interest amongst her peers. Could a general result be proved for *y *= x^{2 }or *y* = x^{4}? What about *y = x*^{n}? Each student chose a particular line of inquiry in the second lesson.

Roisin herself decided to collaborate with Cath who was looking at the most general case (*y *= *mx *and *y* =* x*^{n}). Without assistance from the teacher, they persevered to complete the proof (see Cath's version above).

Unsure of their result, Cath and Roisin invited other students to verify that the particular cases they had generated fitted the final ratio in the proof. Two students claimed that the cases they had worked on were counter-examples. The inquiry ended with the class checking the cases to show that, in fact, they did fit the ratio.

*Andrew Blair, *July 2022

# A surprising result

In a year 12 class, the students asked how it was possible to work out the area under a curve accurately. The teacher explained the concept of integration and its connection to their question. She showed the class how the prompt was true for the equations used in creating the diagram.

The students decided to explore in pairs to find more examples, which they then demonstrated on the board. Two examples are shown below.

Some pairs developed a general approach. For example, two students attempted to derive the relationship between *a*,* b*,* c *and *d* from the equations *y *= *ax*^{3} + *bx*^{2} + *cx* and *y *=* dx*.

One student (Greg Burchett) simplified the equations to the forms *y* = *x*^{n }* *and* y *=* mx*. He then used a **spreads****heet** at home to find the ratio between the area of the triangle and the area under the curve.

When Greg presented his results at the beginning of the next lesson, the class was surprised to see that as *m* varied for a particular value of *n*, the ratio stayed the same. The general form of the ratio, which Greg derived from the pattern he observed in column G of the spreadsheet, is (*n* + 1):2

# Structured inquiry

In another year 12 class, the inquiry started with students speculating that the equations of the straight line and the curve are *y *= 3*x* and *y* = *x*^{3} respectively and that the points of intersection occur at *x* = 0 and* x* = 3. As the discussion developed, one student explained that the second point of intersection must occur at the square root of three if the equations are correct.

However, students were not convinced that the equations were correct and began to justify others. Perhaps the gradient of the straight line was greater than three and perhaps the order of the equation of the curve was also greater than three. The teacher used **Desmos** as students suggested different possibilities, adapting each other's ideas to achieve a better fit. At the end of one chain of reasoning, for example, the students had reached *y *= -*x*^{4} + 3*x*^{3} as the equation of the curve.

When the discussion turned to the area under the curve, students were aware of the trapezium rule. However, they were dissatisfied that it could only give an over- or under-estimation. The class had learnt about the indefinite integral and the teacher now applied their knowledge to the new context.

In the **structured inquiry** that followed the class agreed to restrict the inquiry to the general forms *y *=* mx* and *y *= *x*^{n}. Students explored by substituting in different values for *m* and *n*, finding the points of intersection and integrating to find the ratio of the two areas. The structured questions led the class into more abstract lines of inquiry.

Once the class had become comfortable integrating with upper and lower limits, the teacher explained* why *the definite integral gives the area under the curve.

# Mathematical notes

In the **mathematical notes**, we derive the ratio of

the area under the straight line : the area under the curve

for *y* =* mx *and* y *= *x*^{3}, *y* = 3*x* and *y *= *x*^{n}, and *y *=* mx* and *y* = *x*^{n}.

The statement in the prompt requires students to find the equations of a straight line and a curve that give two areas in the ratio 2:1. In the cases we examine in the notes, the ratio occurs when *n* = 3 for all values of *m*. In general, the ratio is (*n *+ 1):2 for all values of *m*.