Definite Integrals and Area (AI SL)
Once you can find an antiderivative, a definite integral turns it into a number: substitute the upper and lower limits and subtract. That same number is also the area under a curve, provided the curve stays above the \(x\)-axis over the interval. This page covers evaluating definite integrals and using them for area, including where two curves overlap. It's part of the broader Integration topic.
25 questions on this sub-topic.
Evaluating a definite integral
Covered under IB syllabus reference SL5.5: definite integrals using technology, and the area of a region enclosed by a curve \(y=f(x)\) and the \(x\)-axis, where \(f(x)>0\).
Definite integral
\(\displaystyle\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b)-F(a)\)
Find the antiderivative \(F(x)\), then substitute the limits and subtract. No "+C" is needed - it cancels between the two substitutions.
Area under a curve
For \(f(x)>0\) on \([a,b]\), the definite integral gives the area between the curve and the \(x\)-axis directly.
Where a curve dips below the axis, or two curves are involved, the integral needs adjusting before it represents a genuine area (see the worked examples).
Need the full syllabus wording and the antidifferentiation rules behind it? See Integration.
Worked examples
Evaluate \(\displaystyle\int_0^2 (3x^2 - 2x)\,dx.\)
Worked solution
\(\big[x^3 - x^2\big]_0^2.\) M1 A1
\((8 - 4) - 0\) M1 \(= 4.\) A1
Find the area enclosed between \(y = x^2\) and \(y = 2x\).
Worked solution
\(x^2=2x\Rightarrow x=0,2.\) M1
A1
\(\int_0^2(2x-x^2)\,dx=\tfrac43\) M1
\(\approx1.33.\) A1
A1
Evaluate \(\displaystyle\int_0^2 x\,dx.\)
Worked solution
\(\int_0^2 x\,dx\) M1
\(= \left[\dfrac{x^2}{2}\right]_0^2\) A1
\(= 2.\) A1
The curves \(y = x^2\) and \(y = 2x\) meet at \(x = 0\) and \(x = 2.\) Find the area enclosed between them.
Worked solution
Between \(0\) and \(2\), \(2x \ge x^2\), so \(A\) M1 \(= \displaystyle\int_0^2 (2x - x^2)\,dx.\) A1 A1
\(A = \dfrac{4}{3} \approx 1.33.\) M1 A1
Common mistakes
- Treating a negative definite integral as the final area. If a curve dips below the \(x\)-axis, the integral comes out negative - take the modulus to report a sensible (positive) area.
- Forgetting the "+C" on an indefinite integral. Without a boundary condition, \(\displaystyle\int 4x\,dx\) is \(2x^2+C\), not just \(2x^2\) - the constant is part of the answer until you're given a point to solve for it.
- Substituting the limits in the wrong order. A definite integral is \(F(b)-F(a)\), upper limit first - swapping the order flips the sign of the whole answer.
Ready to practise properly?
25 definite-integral and area questions, marked instantly like the real exam.
Quick answers
How do you evaluate a definite integral?
Find the antiderivative \(F(x)\), then compute \(F(b)-F(a)\), where \(a\) and \(b\) are the lower and upper limits. No constant of integration is needed - it cancels out.
Is the area between two curves just their definite integral?
Not directly - first find where the curves intersect to get the limits, then integrate the difference between the upper and lower curve (top minus bottom) between those limits.