Curves that cross the axis

What you will be able to do

• tell the signed area from the total area;
• split an interval at the zeros of a function and add the pieces;
• answer "find the area bounded by the curve and the $x$-axis" when no limits are given.

Below the axis

On a walk across a valley, the height of the ground above a reference level changes from positive to negative. Where the ground is above the level, a strip of the profile has positive height. Where it is below, the height $f(x)$ is negative, so the strip contributes a negative amount to $\int f(x)\,dx$.

So the integral measures signed area: area above the axis minus area below it. Areas themselves are never negative. The total area between the curve and the axis counts the area below the axis as positive:

AP and IB texts call the signed value of the integral the net area. A question that asks for the "area of the region" means the total area.

How to find the total area

1. Solve $f(x) = 0$ to find where the curve meets the $x$-axis.
2. Keep the zeros that lie inside $(a, b)$. They cut $[a, b]$ into pieces.
3. Integrate $f$ over each piece. Take the absolute value of each result.
4. Add the absolute values.

For $f(x) = \sin x$ on $[0, 2\pi]$, the zero inside is $x = \pi$. The piece from $0$ to $\pi$ gives $2$, and the piece from $\pi$ to $2\pi$ gives $-2$. The signed area is $0$ and the total area is $2 + 2 = 4$.

When no limits are given

Often a question says "find the area bounded by the curve and the $x$-axis". The limits are then the points where the curve meets the axis. For $y = x^2 - 4x$ the zeros are $x = 0$ and $x = 4$. Between them the curve is below the axis, and $\displaystyle\int_0^4 (x^2 - 4x)\,dx = -\dfrac{32}{3}$. The area is $\dfrac{32}{3}$.

In an AP or IB question, "the integral of $f$" is the signed value, and "the area" is the total. An Indian question says "the area bounded by", which is also the total.

A worked example

Find the signed area and the total area of $y = x(x - 2)(x - 4)$ from $x = 0$ to $x = 4$.

1. The zeros. They are $x = 0$, $2$ and $4$. The only zero inside the interval is $x = 2$.
2. Expand. $x(x - 2)(x - 4) = x^3 - 6x^2 + 8x$, with antiderivative $\dfrac{x^4}{4} - 2x^3 + 4x^2$.
3. The first piece. From $0$ to $2$ the value is $4 - 16 + 16 = 4$. The curve is above the axis.
4. The second piece. From $2$ to $4$ the value is $0 - 4 = -4$. The curve is below the axis.
5. Combine. The signed area is $4 + (-4) = 0$. The total area is $4 + 4 = 8$.

A second quick case is the line $y = x - 2$ on $[0, 5]$. The piece from $0$ to $2$ gives $-2$, and the piece from $2$ to $5$ gives $4.5$. The signed area is $2.5$ and the total area is $6.5$.

Practise with new numbers

Try it in Python

The program cuts $[0, 4]$ into $4000$ thin strips for the curve of the worked example. It adds the heights with their signs and also adds their absolute values.

Common mistakes

• Integrating across a zero. The two parts cancel, and the answer is too small. Split at every zero inside the interval.
• Forgetting the absolute value. A piece below the axis gives a negative number. Change its sign before you add.
• Using the zeros outside the interval. Only the zeros between $a$ and $b$ cut the interval.
• Reporting a negative area. An area is never negative. A negative result means that the region lies below the axis, so change its sign.

Remember these

• The integral is the signed area. The total area is $\displaystyle\int_a^b \lvert f(x) \rvert\,dx$.
• Split the interval at every zero of $f$ inside it, integrate each piece and add the absolute values.
• With no limits given, the limits are the points where the curve meets the axis.
• A signed area of $0$ does not mean that there is no region.

Check yourself

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

1. Find the signed area and the total area of $y = \cos x$ from $x = 0$ to $x = \pi$.

2. Find the area bounded by the curve $y = 4 - x^2$ and the $x$-axis.

3. Find the signed area and the total area of $y = 2x - 6$ from $x = 0$ to $x = 6$.

4. Why is $\displaystyle\int_0^{2\pi} \sin x\,dx$ not the area of the region between one full wave of the sine curve and the axis?

5. The curve $y = x(x - 2)(x - 4)$ is drawn from $x = -1$ to $x = 5$. Into how many pieces must the interval be cut to find the total area?

References

• OpenStax, Calculus Volume 1, Chapter 5, section on the definite integral (net signed area and total area).
• NCERT, Mathematics Part II for Class XII, Chapter 8, section on the area under simple curves.