Negative Area Below the x-axis

Q: Why is my area coming out negative?

Because the region lies below the x-axis, where y<0y<0y<0, so ∫y dx\int y\,dx∫ydx is negative. The definite integral gives signed area. Take the absolute value of the negative result to get the physical area.

Q: What if the curve is partly above and partly below the axis?

Split the integral at the x-intercepts. Integrate each part separately, take the absolute value of any piece that is below the axis, then add the sizes. Integrating straight across in one go lets the positive and negative parts cancel and understates the true area.

Q: Can I just make the answer positive at the end?

Only when the entire region is on one side of the axis. If it crosses the axis, blindly taking the absolute value of a single integral is wrong, because cancellation has already happened inside it. Split first, then take sizes.

Q: Do I always write units²?

Yes, area is measured in square units. If the axes carry no specific unit, write units². Leaving the units off can cost a mark even when the number is right.

The error

When a region is bounded by a curve and the x-axis, it is tempting to reach straight for $\int_a^b y\,dx$ and report whatever number comes out as the area. That works only while the curve stays above the axis.

Where the curve dips below the x-axis, $y$ is negative, so the integral over that stretch is negative too, the definite integral measures signed area, not physical area. Students then write a negative area, which cannot exist, or they cross out the minus sign without understanding why.

The slip is natural: for every region above the axis the integral and the area are the same number, so the habit of quoting the integral directly seems to work, right up until the region drops below the axis and the answer turns negative on you.

Wrong line, then the fix

Take the region enclosed by $y=x^{2}-4$ , the x-axis, and the lines $x=0$ and $x=2$ . On this interval the curve lies entirely below the axis (for example $y=-4$ at $x=0$ ), so the integral must come out negative.

Signed value of the integral, correct, but this is not the areaMust memorise

\int_{0}^{2}(x^{2}-4)\,dx=\left[\frac{x^{3}}{3}-4x\right]_{0}^{2}=\frac{8}{3}-8=-\frac{16}{3}

Reporting the area as $-\frac{16}{3}$ is the mistake: area is a size and cannot be negative. The integral is telling you the region lies below the axis; take its magnitude.

Correct areaMust memorise

\text{Area}=\left|\int_{0}^{2}(x^{2}-4)\,dx\right|=\frac{16}{3}\text{ units}^{2}

The reliable fix

Find where the curve meets the x-axis by solving $y=0$ ; these roots may split your interval.
On each sub-interval, test one point to see whether the curve is above ( $y>0$ ) or below ( $y<0$ ) the axis.
Integrate over each sub-interval separately.
For any piece below the axis, take the absolute value of its (negative) integral.
Add the sizes of all pieces to get the total area, and finish with units².

One line to remember it

Area is a size, never a sign

If your definite integral comes out negative, the region is below the x-axis, keep the number, drop the minus sign, and write units².

Practice where the error hides

Q1[3 marks]

The curve $y=x^{2}-2x$ meets the x-axis at $x=0$ and $x=2$ . Find the area of the region enclosed between the curve and the x-axis.

Show worked solution

First check the sign of $y$ between the roots. At $x=1$ , $y=1-2=-1<0$ , so the whole region lies below the x-axis; the integral will be negative.

\int_{0}^{2}(x^{2}-2x)\,dx=\left[\frac{x^{3}}{3}-x^{2}\right]_{0}^{2}=\left(\frac{8}{3}-4\right)-0=-\frac{4}{3}

The integral is $-\frac{4}{3}$ ; because the region is below the axis, the area is its magnitude.

\text{Area}=\left|-\frac{4}{3}\right|=\frac{4}{3}\text{ units}^{2}

The area is $\frac{4}{3}$ units². Quoting $-\frac{4}{3}$ would lose the accuracy mark, even though the integration itself was correct.

How one-to-one teaching helps

Our teachers train the habit of sketching the curve first, so you can see at a glance which parts of a region sit below the x-axis before you integrate anything. In a one-to-one lesson we pause at the exact line where a negative answer appears and ask the question that fixes it for good, is this a size, or a signed value?

Because the paper awards method marks, showing the split at the roots and the absolute-value step keeps your marks safe even if the arithmetic wobbles. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

Message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

Why is my area coming out negative?

Because the region lies below the x-axis, where $y<0$ , so $\int y\,dx$ is negative. The definite integral gives signed area.

Take the absolute value of the negative result to get the physical area.

What if the curve is partly above and partly below the axis?

Split the integral at the x-intercepts. Integrate each part separately, take the absolute value of any piece that is below the axis, then add the sizes.

Integrating straight across in one go lets the positive and negative parts cancel and understates the true area.

Can I just make the answer positive at the end?

Only when the entire region is on one side of the axis. If it crosses the axis, blindly taking the absolute value of a single integral is wrong, because cancellation has already happened inside it.

Split first, then take sizes.

Do I always write units²?

Yes, area is measured in square units. If the axes carry no specific unit, write units².

Leaving the units off can cost a mark even when the number is right.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Area below the x-axis coming out negative

The error

Wrong line, then the fix

The reliable fix

One line to remember it

Practice where the error hides

How one-to-one teaching helps

Frequently asked questions

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