Area Between Two Curves Setup

Q: Do I subtract the two areas or integrate the difference?

Integrate the difference. Write one integral of (upper curve − lower curve) between the crossing points. Working out each area separately and subtracting only agrees with this when both curves stay above the x-axis across the whole strip.

Q: Which curve goes first in the bracket?

The upper one, the curve with the larger yyy between the limits. Test any x-value between the crossings and substitute into both curves to see which is on top. Putting the lower curve first gives a negative answer.

Q: Where do the limits come from?

From the points where the two curves meet. Set their yyy expressions equal and solve for xxx. Those x-values are the limits; do not use other numbers from the question unless the region is also cut off by a stated vertical line.

The error

Faced with the region trapped between two curves, students often reach for one of three broken set-ups. Some integrate each curve down to the x-axis separately and subtract the two areas, which only behaves if both curves stay above the axis across the whole strip.

Some subtract in the wrong order, writing (lower − upper), and are then puzzled by a negative area. Some never solve for the intersection points and instead borrow limits from the question that were never the crossing points at all.

Each slip is tempting because the individual pieces, integrate a curve, subtract two numbers, are all familiar. The trick the exam is testing is combining them into one clean integral: the height of the region at any $x$ is the top curve minus the bottom curve, and that single difference is what you integrate between the crossings.

Wrong line, then the fix

Take the region enclosed by the line $y=x+2$ and the parabola $y=x^{2}$ . They meet where $x+2=x^{2}$ , that is $x^{2}-x-2=0$ , so $(x-2)(x+1)=0$ and the curves cross at $x=-1$ and $x=2$ .

Testing $x=0$ gives $y=2$ on the line and $y=0$ on the parabola, so the line is the upper curve on this interval.

Wrong, lower curve minus upper curve gives a negative areaMust memorise

\text{Area}=\int_{-1}^{2}\bigl[x^{2}-(x+2)\bigr]\,dx=-\frac{9}{2}

A negative area is impossible; the minus sign is the signal that the subtraction ran the wrong way. Swap to (upper − lower) and integrate once between the crossing points.

Correct set-up: upper curve minus lower curveMust memorise

\text{Area}=\int_{-1}^{2}\bigl[(x+2)-x^{2}\bigr]\,dx=\left[\frac{x^{2}}{2}+2x-\frac{x^{3}}{3}\right]_{-1}^{2}

At $x=2$ the bracket is $2+4-\tfrac{8}{3}=\tfrac{10}{3}$ ; at $x=-1$ it is $\tfrac{1}{2}-2+\tfrac{1}{3}=-\tfrac{7}{6}$ . Subtracting the lower limit from the upper limit:

Correct areaMust memorise

\text{Area}=\frac{10}{3}-\left(-\frac{7}{6}\right)=\frac{20}{6}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}\text{ units}^{2}

The reliable fix

1
Find where the curves meet
Set the two expressions for $y$ equal and solve. These x-values are your limits of integration, do not borrow numbers from elsewhere in the question.
2
Decide which curve is on top
Pick any x-value between the crossings and substitute into both curves. The larger $y$ is the upper curve for the whole strip.
3
Write one integral of (upper − lower)
The height of the region at each $x$ is the top curve minus the bottom curve. Integrate that single difference between the limits.
4
Evaluate carefully
Substitute the upper limit, then the lower limit, and subtract. A correct set-up returns a positive number.
5
Sanity-check the sign
If the answer is negative, you subtracted the wrong way round, swap upper and lower, then finish with units².

One line to remember it

Top minus bottom, between the crossings

One integral: upper curve minus lower curve, taken between the x-values where they meet. If it comes out negative, you flipped the order.

Practice where the error hides

Q1[4 marks]

Find the area of the region enclosed by the curve $y=x^{2}$ and the line $y=4$ .

Show worked solution

First find where the curve meets the line by setting the expressions equal.

x^{2}=4\;\Rightarrow\; x=\pm 2

So the limits are $x=-2$ and $x=2$ . Between them, test $x=0$ : the line gives $y=4$ and the curve gives $y=0$ , so the line $y=4$ is the upper boundary.

Writing it the wrong way as $\int_{-2}^{2}(x^{2}-4)\,dx$ would give $-\tfrac{32}{3}$ , a negative 'area' that flags the reversed order. Set it up as (upper − lower):

\text{Area}=\int_{-2}^{2}\bigl[4-x^{2}\bigr]\,dx=\left[4x-\frac{x^{3}}{3}\right]_{-2}^{2}

At $x=2$ : $8-\tfrac{8}{3}=\tfrac{16}{3}$ . At $x=-2$ : $-8+\tfrac{8}{3}=-\tfrac{16}{3}$ .

\text{Area}=\frac{16}{3}-\left(-\frac{16}{3}\right)=\frac{32}{3}\text{ units}^{2}

The area is $\frac{32}{3}$ units². The intersection points supplied the limits, the point test fixed which curve was on top, and one integral of the difference did the rest.

How one-to-one teaching helps

Our teachers turn this into a fixed opening routine: solve for the crossings, mark which curve is higher, then write a single integral of (upper − lower). In a one-to-one lesson we watch the exact line where the set-up goes wrong, a borrowed limit or a flipped subtraction, and correct it before it hardens into a habit.

Because the paper awards method marks, a clearly shown difference and correct limits protect your marks even if the arithmetic slips. 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

Do I subtract the two areas or integrate the difference?

Integrate the difference. Write one integral of (upper curve − lower curve) between the crossing points.

Working out each area separately and subtracting only agrees with this when both curves stay above the x-axis across the whole strip.

Which curve goes first in the bracket?

The upper one, the curve with the larger $y$ between the limits. Test any x-value between the crossings and substitute into both curves to see which is on top.

Putting the lower curve first gives a negative answer.

Where do the limits come from?

From the points where the two curves meet. Set their $y$ expressions equal and solve for $x$ .

Those x-values are the limits; do not use other numbers from the question unless the region is also cut off by a stated vertical line.

My answer is negative, what went wrong?

You almost certainly subtracted lower minus upper. Area is a size and cannot be negative, so swap the order to (upper − lower) and integrate again.

The magnitude is usually right; only the sign was reversed.

Source:SRC-DSKP-EN

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

Setting up the area between two curves wrongly

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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