Definite Integrals and Area (AI HL)
A definite integral turns an antiderivative into a number: substitute the top limit, subtract the value at the bottom limit. This page covers evaluating these directly and using them to find the area under a curve. It's part of the broader Integration topic.
10 questions on this sub-topic.
The two formulas
Covered under IB syllabus reference SL5.5: "Definite integrals using technology. Area of a region enclosed by a curve \(y=f(x)\) and the \(x\)-axis, where \(f(x)>0\)." Both formulas below are in the booklet.
Standard integrals
\[\int e^x\,dx=e^x+c,\quad \int\dfrac1x\,dx=\ln|x|+c\]
Memorise these alongside \(\sin x\), \(\cos x\) and \(\dfrac{1}{\cos^2x}\) - they appear constantly in AI HL papers.
✓ In the formula bookletArea under a curve
\[A=\int_a^b f(x)\,dx\]
Only gives the true area directly when \(f(x)>0\) throughout \([a,b]\) - otherwise split at the roots first.
✓ In the formula bookletFor entering a definite integral on your GDC, see the parent Integration page.
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
The average value of \(f\) on \([a,b]\) is \(\dfrac{1}{b-a}\int_a^b f(x)\,dx.\)
Find the average value of \(f(x)=x^2\) on \([0,3]\).
Worked solution
\(\tfrac13\int_0^3 x^2\,dx\) M1
\(= \tfrac13[\tfrac{x^3}{3}]_0^3 = \tfrac13\cdot 9 = 3.\) A1
Evaluate \(\displaystyle\int_0^{\ln 3} \dfrac{e^x}{e^x+1}\,dx.\)
Worked solution
The numerator is the derivative of the denominator, so an antiderivative is \(\ln(e^x+1).\) M1
\(\big[\ln(e^x+1)\big]_0^{\ln 3}.\) A1
\(\ln(e^{\ln 3}+1)=\ln(3+1)=\ln 4.\) A1
\(\ln(e^0+1)=\ln 2,\) so the integral \(=\ln4-\ln2=\ln2.\) A1
\(\approx0.693.\) A1
Evaluate \(\displaystyle\int_1^2 e^{2x}\,dx,\) giving your answer to 3 significant figures.
Worked solution
\(\Big[\dfrac{1}{2}e^{2x}\Big]_1^2.\) M1 A1
\(\dfrac{1}{2}e^4-\dfrac{1}{2}e^2\) M1 \(=23.6045\ldots\approx 23.6.\) A1
Common mistakes
- Treating a definite integral as area without checking the sign. If the curve dips below the \(x\)-axis anywhere in the interval, the raw integral gives a signed (net) value, not the true area - split at the roots and take magnitudes first.
- Only substituting the top limit. A definite integral needs \(F(b) - F(a)\), not just \(F(b)\) - forgetting to subtract the lower-limit value is an easy way to lose the final mark.
- Subtracting the curves the wrong way round when the region borders another curve rather than the axis. Area needs (top \(-\) bottom); getting the order backwards gives a negative answer that should have been positive.
Ready to practise properly?
11 definite-integral questions, marked instantly like the real exam.
Quick answers
How do you evaluate a definite integral?
Find the antiderivative \(F(x)\), then substitute the upper limit and subtract the value at the lower limit: \(\int_a^b f(x)\,dx = F(b) - F(a).\)
Does a definite integral always give the area?
Only when the function stays positive across the whole interval. If it dips below the \(x\)-axis, the raw integral gives a signed value, so split at any roots and take magnitudes to get the true area.