Finding the area under a curve

A definite integral measures the area of the region trapped between a curve and the $x$-axis, over a chosen interval. This method turns that geometric idea, the area of a curved region you could not find with simple shapes, into a calculation you can do exactly with integration.

In Add Math, this is the headline application of the Integration chapter. A typical question gives a curve $y=f(x)$ and two vertical boundaries $x=a$ and $x=b$, and asks for the area enclosed by the curve, the $x$-axis and those lines.

The area is the definite integral $\int_{a}^{b}y\,dx$. You integrate the function, substitute the upper and lower limits, and subtract.

Because it builds directly on integrating polynomials, a secure grasp of the power rule makes this method quick and reliable.

Reach for this method whenever a question asks for the area of a region bounded by a curve, the $x$-axis, and two vertical lines, or asks for 'the area enclosed by the curve and the $x$-axis between $x=a$ and $x=b$'. A definite integral with limits already written is the same signal.

Check first whether the curve stays above the $x$-axis across the interval. If it does, the integral gives the area directly.

If part of the region lies below the axis, the integral there comes out negative, so you split the interval at the point where the curve crosses the axis and take the size of each piece. When the region is bounded by the $y$-axis instead, you integrate with respect to $y$, but the everyday case in this chapter is area under a curve above the $x$-axis.

1. 1

   Identify the region

   Note the curve, the $x$-axis, and the two limits $x=a$ and $x=b$; a quick sketch helps.

2. 2

   Set up the integral

   Write the area as $\int_{a}^{b}y\,dx$ with the correct limits.

3. 3

   Integrate

   Integrate $y$ term by term using the power rule; do not add $c$ for a definite integral.

4. 4

   Substitute the limits

   Put the upper limit $b$ and the lower limit $a$ into the integrated expression.

5. 5

   Subtract

   Compute $F(b)-F(a)$.

6. 6

   State the area

   Give the answer as a positive area, with square units.

• Adding $+c$ to a definite integral. The constant cancels in $F(b)-F(a)$, so it is not written.
• Substituting the lower limit first, or subtracting the wrong way round: it is always $F(b)-F(a)$.
• Assuming a negative answer is impossible and ignoring a sign; a region below the $x$-axis gives a negative integral, so split and take the size of each part.
• Substituting the limits into $y$ instead of into the integrated expression $F(x)$.
• Forgetting the units, an area is given in square units.

The step that quietly costs marks is the boundaries: setting the right limits and knowing whether the region dips below the $x$-axis. In a one-to-one lesson our teachers start every area question with a quick sketch, so you can see the region before you integrate and decide whether to split it.

We also make the square-bracket notation and the $F(b)-F(a)$ subtraction a fixed routine, which is where careless sign errors usually creep in. Because the paper awards method marks, a clearly set-up integral scores even if a number slips.

Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English. To practise area questions with clear sketches, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.