Mathematics, four units · Grade 11

How do you find the area enclosed between two graphs?

First find where they intersect, because those points are the limits of integration. Then evaluate the definite integral of the upper function minus the lower one between those limits. The order of the subtraction is the whole point: whichever curve lies above on that interval is written first, otherwise the area comes out negative.

Learning objectives

The antiderivative

Integration is the reverse of differentiation: you look for a function whose derivative is the one you were given. For a polynomial the rule is simple — raise each power by one and divide by the new power.

Every function has infinitely many antiderivatives, differing by a constant. That is why an indefinite integral always carries a plus C. Dropping the constant is a mistake that costs a mark.

In a definite integral the constant cancels in the subtraction, so there it can be ignored. That is the only reason it may be left out — not because it does not matter.

Area between a curve and the axis

The area enclosed between a graph and the horizontal axis is the definite integral between the points where the graph meets that axis. Find those first by setting the function to zero.

If the graph lies below the axis on part of the range, the integral comes out negative. Area is not negative, so take the absolute value — or compute in the reverse order.

When the graph crosses the axis inside the range, split into two separate integrals. A single integral across the crossing cancels positive area against negative and gives a number that is not the area.

Area between two graphs

For the area between two curves, the limits are their points of intersection. Set the two functions equal to each other and solve.

Inside the interval, identify which function lies above. The safe way is to pick a single point inside and substitute into both — the larger value belongs to the upper curve.

Then integrate the upper minus the lower. This formula also works when part of the region lies below the horizontal axis, which is why it is preferable to computing two areas separately.

Worked examples

  1. Find the indefinite integral of 6x squared minus 4x.

    1. First term: raise the power to 3 and divide by 3, giving 2x cubed
    2. Second term: raise to 2 and divide by 2, giving minus 2x squared
    3. Add the constant

    Answer: 2x cubed minus 2x squared plus C

  2. What is the area between f(x) equals x squared and the x-axis, from 0 to 3?

    1. The antiderivative is x cubed over 3
    2. At 3: 27 over 3, which is 9
    3. At 0: zero
    4. Subtract

    Answer: 9

  3. What is the area enclosed between y equals x plus 2 and y equals x squared?

    1. Intersections: x plus 2 equals x squared, so x squared minus x minus 2 equals zero
    2. The solutions are minus 1 and 2
    3. Test at x equals 0: the line gives 2 and the parabola gives zero, so the line is above
    4. Integrate (x plus 2 minus x squared) from minus 1 to 2

    Answer: 4.5

Common mistakes

Forgetting the constant in an indefinite integral
All the functions differing by a constant share the same derivative, so the answer is a family rather than a single function. In a definite integral the constant cancels, and only there may it be omitted.
Subtracting the wrong way round and getting a negative area
The upper function always comes first. A negative number in an area answer is a certain sign that the order was swapped, not a reason to slap on an absolute value without checking.
Using one integral across a range where the graph crosses the axis
The part below the axis counts as negative and cancels the part above. Split at the crossing point and add the absolute values, or the area comes out too small or even zero.

What to remember

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