This tool takes a polynomial and two bounds, then shows the antiderivative, the signed definite integral and the total area. It is for Form 4 and Form 5 students who can integrate but get confused when an area comes out as zero or negative.
Free tool · Additional Mathematics
Polynomial integration and area explorer
You can integrate, but an area that comes out as zero or negative looks wrong.
Everything you enter stays on this device.
Who is this tool for?
It helps a student who has just met definite integrals in Additional Mathematics. It also suits a student revising for SPM who wants to check a practice answer by looking at the shaded region.
How do I use it?
- Type the coefficients of f(x). For example, f(x) = x has a coefficient of 1 for x and 0 for everything else.
- Enter the lower limit a and the upper limit b, in the order the question gives them.
- Press “Work it out”.
- Read the antiderivative, then the signed integral, then the total area and the table of pieces.
- Look at the graph. Green is above the x-axis and red is below.
How does it work?
The tool raises each power of x by 1 and divides by the new power to get F(x) + C. The signed integral is F(b) − F(a), and the constant C cancels.
To find the area, the tool searches for the points where f(x) crosses the x-axis between the bounds. It splits the interval there and adds the size of each piece. Crossing points are found numerically and shown to 4 decimal places.
What does a worked example look like?
Take f(x) = x² − 4 from a = 0 to b = 3. The antiderivative is x³/3 − 4x + C.
F(3) = 9 − 12 = −3 and F(0) = 0, so the signed integral is −3. The curve crosses the x-axis at x = 2. From 0 to 2 the piece is −16/3, and from 2 to 3 it is 7/3. The area is 16/3 + 7/3 = 23/3.
The tool shows decimals, so the pieces appear as −5.333333 and 2.333333 and the area as 7.666667.
The signed integral is −3 but the area is about 7.6667. Writing −3 as an area would be a mistake, and the tool points this out.
What can the tool not tell you?
It does not find the area between two curves. It does not handle functions that are not polynomials, and it does not show your working in the form an exam expects.
Use it to check and understand, then write the full method yourself. Read more in evaluating definite integrals.
Where does this connect to the lessons?
The antiderivative and the constant are taught in finding an indefinite integral and its constant. The link between integrals and motion is in displacement and distance.
When you are ready for questions, try the integration practice. If you are unsure whether a question needs a derivative or an integral, read choosing between a derivative and an integral.
For lessons with a teacher on your own questions, see Additional Mathematics tuition and the one-hour trial class (from RM50).
Common questions
Why is the integral 0 but the area is not 0?
The integral is signed. Parts of the curve above the x-axis count as positive and parts below count as negative, so they can cancel to zero. Area counts every part as positive, so it adds the sizes of the pieces instead.
What happens if I swap the two bounds?
The signed integral changes sign, because going from b to a is the reverse of going from a to b. The area stays the same, since the region between the curve and the x-axis has not moved.
Does the tool work for any function?
It handles polynomials up to degree 4 with real coefficients and finite bounds. Other functions, such as fractions with x in the denominator, are outside its limits and are not supported.
If the tool shows a gap between the signed integral and the area, a teacher can work through that step with you on your own questions in a one-to-one lesson.
- Online one-to-one lessons for your child with an experienced teacher.
- Your first class is a one-hour trial, from RM50. The fee is agreed before you book.
- Happy with the teacher? Continue with lessons of about 1.5 hours. If not, ask for another teacher.