Definite Integrals and Area (AA HL)
A definite integral turns an antiderivative into an actual number: evaluate it at the upper limit, evaluate it at the lower limit, and subtract. That single idea is also how you find the area trapped between a curve and the \(x\)-axis, or between two curves. This page covers the technique, the sign traps that come with it, and the mistakes that cost the most marks. It's part of the broader Integration topic.
26 questions on this sub-topic.
Key ideas
Covered under IB syllabus reference SL5.11: definite integrals via \(\int_a^b g'(x)\,dx = g(b)-g(a)\), and the areas of regions bounded by a curve and the \(x\)-axis or between two curves.
Standard integrals
\(\displaystyle\int \sin x\,dx=-\cos x+C,\ \int\cos x\,dx=\sin x+C,\ \int e^x\,dx=e^x+C\)
In the formula booklet. You'll usually need one of these before you can even apply the limits.
Area below the axis
Where a curve dips below the \(x\)-axis, its definite integral over that stretch comes out negative. Take the modulus of that piece before adding it to the total area.
Not in the formula booklet - reasoning stepAreas between curves
Integrate the difference of the two functions, top minus bottom, between their points of intersection.
Not in the formula booklet - techniqueNeed the full syllabus wording and formula-booklet reference table, or a GDC walkthrough for numerical integration? See Integration.
Worked examples
Evaluate \(\displaystyle\int_0^{\pi} \sin x\,dx.\)
Worked solution
\([-\cos x]_0^{\pi}\) M1
\(-\cos x\) A1
\(= -\cos\pi + \cos 0 = 1 + 1 = 2.\) A1
The region under \(y=\sqrt x\) from \(x=0\) to \(x=4\) is rotated about the \(x\)-axis.
Find the volume.
Worked solution
for rotation about the \(x\)-axis: \(V = \pi\int_a^b y^2\,dx.\) M1
\(y = \sqrt x\) so \(y^2 = x.\) A1
\(V = \pi\int_0^4 x\,dx\) M1 \(= \pi\left[\tfrac{x^2}{2}\right]_0^4.\) A1
\(V = 8\pi \approx 25.1.\) M1 A1
This is a calculator paper - you could also evaluate \(\pi\int_0^4 x\,dx\) directly with your GDC's numerical integral function. See Integration for the keystrokes.
Common mistakes
- Not taking the modulus when a curve dips below the axis. A definite integral over a region where \(f(x)<0\) comes out negative - if the question asks for an area, take the modulus of that piece before adding it to the rest.
- Forgetting "+C" on an indefinite integral. Every indefinite integral represents a family of curves, not a single one - dropping the constant loses an accuracy mark even if the rest of the working is correct.
- Subtracting the curves the wrong way round. When finding the area between two curves, always integrate top minus bottom between their intersection points - if you integrate bottom minus top by mistake, you'll get a negative value instead of the true area.
Ready to practise properly?
29 definite-integral and area questions, marked instantly like the real exam.
Quick answers
How do I find the area under a curve using integration?
Evaluate the definite integral of the function between the two \(x\)-values that bound the region. If any part of the curve dips below the \(x\)-axis, that piece of the integral comes out negative, so take its modulus before adding it to the rest.
What is the difference between a definite and an indefinite integral?
An indefinite integral has no limits and gives a family of antiderivatives plus a constant \(C\). A definite integral has limits \(a\) and \(b\) and evaluates to an actual number (or expression), with no constant of integration needed.