Definite Integrals and Area (AA SL)
A definite integral has limits, so instead of a family of curves it gives you a single number - and once you have that number, you can read it as the signed area between a curve and the \(x\)-axis. This page covers evaluating definite integrals with the fundamental theorem and using them to find enclosed areas, with worked examples and the mistakes that catch people out. It's part of the broader Integration topic.
21 questions on this sub-topic.
Fundamental theorem and area
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 \(f(x)>0\). The fundamental theorem is in the formula booklet; treating an integral as an area is your own application of it, so it isn't listed separately.
Fundamental theorem
\(\displaystyle\int_a^b g'(x)\,dx = g(b)-g(a)\)
Find the antiderivative, then substitute the upper limit minus the lower limit. No \(+C\) needed - it cancels.
Area under a curve
Area \(=\displaystyle\int_a^b f(x)\,dx\), for \(f(x)\ge0\) on \([a,b]\)
Not a separate booklet formula - it's the definite integral itself, interpreted as area. For area between two curves, integrate (top minus bottom).
Need the full syllabus wording and formula-booklet reference table? See Integration.
Worked examples
Evaluate \(\displaystyle\int_1^3 (2x + 1)\,dx\).
Worked solution
\(\int(2x+1)\,dx=x^2+x.\) M1 A1
\([x^2+x]_1^3=(9+3)-(1+1)\) M1 \(=10.\) A1
Find the area enclosed between \(y = x^2\) and \(y = 2x.\)
Worked solution
\(x^2 = 2x \Rightarrow x\) M1 \(= 0, 2.\) A1
is \(y = 2x\) on \([0,2].\) M1
\(\int_0^2 (2x - x^2)\,dx\) A1 \(= \left[x^2 - \tfrac{x^3}{3}\right]_0^2\) A1 \(= \tfrac43 \approx 1.33.\) A1
On the GDC: evaluate \(\int_0^2 (2x-x^2)\,dx\) numerically - TI-84 Plus CE via MATH → fnInt, Casio fx-CG50/CG100 via Run-Matrix → ∫dx, TI-Nspire CX via menu → Calculus → Numerical Integral.
Evaluate \(\displaystyle\int_0^{\pi} \sin x \, dx\).
Worked solution
\(\int\sin x\,dx=-\cos x.\) M1A1
\([-\cos x]_0^{\pi}=-\cos\pi-(-\cos 0)=1+1=2.\) M1A1
Common mistakes
- Treating a negative definite integral as a negative area. Where the curve dips below the \(x\)-axis the integral comes out negative - area itself is always positive, so take the absolute value.
- Forgetting "\(+C\)" on an indefinite integral. Even when the question doesn't obviously need it, the accuracy mark is often tied to including the constant - drop it and you lose the mark.
- Integrating (bottom minus top) for an area between two curves. Check which curve lies above the other over the interval first - integrating in the wrong order gives a negative answer for what should be a positive area.
Ready to practise properly?
22 definite-integral and area questions, marked instantly like the real exam.
Quick answers
How do I evaluate a definite integral?
Find the antiderivative, then apply \(\int_a^b g'(x)\,dx = g(b)-g(a)\). No \(+C\) is needed since it cancels. See the parent Integration page for GDC guidance on evaluating numerically.
How do I find the area enclosed between a curve and the x-axis?
Area \(=\int_a^b f(x)\,dx\), provided \(f(x)\ge0\) on \([a,b]\). Where the curve dips below the axis, take the absolute value of that portion before adding it.