Definite integrals and the modulus function
Work with definite integrals and the modulus function in Year 12 Specialist Mathematics for Queensland (QCAA). Evaluate integrals with the fundamental theorem of calculus and use the properties of the definite integral — linearity, reversed limits, adjacent intervals and odd-function symmetry — to simplify the working.
You will learn to tell a signed integral apart from the total area between a curve and the axis, and to integrate \(|f(x)|\) by splitting the domain at the zeros of \(f\) — a core Unit 4 integration technique that sets up areas, volumes and further calculus.
Theory
Definite integrals and the modulus function bring together the fundamental theorem of calculus and the algebra of the definite integral in Year 12 Specialist Mathematics (QCAA, Queensland). You evaluate integrals with known anti-derivatives, use the properties of definite integrals to simplify them, and find the total area between a curve and the axis by integrating \(|f(x)|\) — splitting the domain at the zeros of \(f\).
A definite integral \(\displaystyle\int_a^b f(x)\,dx\) is evaluated with the fundamental theorem of calculus: if \(F\) is any anti-derivative of \(f\), then \(\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)\). Its value is a signed quantity — area above the \(x\)-axis counts as positive and area below counts as negative.
The properties of definite integrals let you rework an integral before evaluating it. It is linear, so constants come out and sums split; reversing the limits changes the sign; an integral over a point is zero; and an integral splits over adjacent intervals. A useful special case: the integral of an odd function over an interval symmetric about \(0\) is \(0\).
The signed integral is not the same as the total area between the curve and the axis. Where the curve dips below the axis the integral subtracts that region, so a signed integral can be small, or even zero, while the actual area is not.
To find the total area you integrate the modulus function \(|f(x)|\). Because \(|f(x)|\) reflects every below-axis part of the graph upward, you split the domain at the zeros of \(f\), integrate \(f\) over each piece, and add the absolute values. The standard anti-derivative \(\displaystyle\int\dfrac{1}{x}\,dx=\ln|x|+c\) uses the modulus for the same reason: it keeps the argument positive.
Evaluate with the fundamental theorem of calculus, where \(F'=f\):
The properties of the definite integral:
For an odd function on a symmetric interval:
The total area between \(y=f(x)\) and the \(x\)-axis on \([a,b]\), where \(f\) has zeros \(c_1
Integrating the modulus function \(|f(x)|\)
- Find the zeros: solve \(f(x)=0\) to locate every point in \([a,b]\) where the graph crosses the \(x\)-axis.
- Split the domain at those zeros into sub-intervals on which \(f\) keeps one sign.
- Integrate each piece of \(f\); take the absolute value of any piece where \(f<0\) (equivalently, integrate \(-f\) there).
- Add the absolute areas to get the total area, and compare with the signed integral if both are asked for.
Use linearity — constants come out and the integral splits:
| \(\int_0^2\big(3f-2g\big)\,dx\) | \(=\) | \(3\int_0^2 f\,dx-2\int_0^2 g\,dx\) |
| \(=\) | \(3(5)-2(3)\) | |
| \(=\) | \(15-6\) | |
| \(=\) | \(9\) |
\(\displaystyle\int_0^2\big(3f(x)-2g(x)\big)\,dx=9\).
Split \([1,4]\) at \(x=2\), then make the wanted piece the subject:
| \(\int_1^4 f\,dx\) | \(=\) | \(\int_1^2 f\,dx+\int_2^4 f\,dx\) |
| \(10\) | \(=\) | \(3+\int_2^4 f\,dx\) |
| \(\int_2^4 f\,dx\) | \(=\) | \(7\) |
Reversing the limits changes the sign:
| \(\int_4^2 f\,dx\) | \(=\) | \(-\int_2^4 f\,dx\) |
| \(=\) | \(-7\) |
\(\displaystyle\int_2^4 f(x)\,dx=7\) and \(\displaystyle\int_4^2 f(x)\,dx=-7\).
The corner is at \(x=1\); split there. For \(x<1\), \(|x-1|=1-x\):
| \(A\) | \(=\) | \(\int_0^1 (1-x)\,dx+\int_1^4 (x-1)\,dx\) |
Integrate each piece:
| \(\int_0^1 (1-x)\,dx\) | \(=\) | \(\left[x-\dfrac{x^2}{2}\right]_0^1\) |
| \(=\) | \(1-\dfrac{1}{2}=\dfrac{1}{2}\) | |
| \(\int_1^4 (x-1)\,dx\) | \(=\) | \(\left[\dfrac{x^2}{2}-x\right]_1^4\) |
| \(=\) | \((8-4)-\left(\dfrac{1}{2}-1\right)=\dfrac{9}{2}\) |
Add the two absolute areas:
| \(A\) | \(=\) | \(\dfrac{1}{2}+\dfrac{9}{2}\) |
| \(=\) | \(5\) |
The total area is \(5\) square units.
Signed integral first — a single anti-derivative over \([0,3]\):
| \(\int_0^3 (x^2-1)\,dx\) | \(=\) | \(\left[\dfrac{x^3}{3}-x\right]_0^3\) |
| \(=\) | \((9-3)-0=6\) |
For the area, \(x^2-1=0\) at \(x=1\); split there (below on \([0,1]\)):
| \(A\) | \(=\) | \(\int_0^1 (1-x^2)\,dx+\int_1^3 (x^2-1)\,dx\) |
| \(\int_0^1 (1-x^2)\,dx\) | \(=\) | \(\left[x-\dfrac{x^3}{3}\right]_0^1=\dfrac{2}{3}\) |
| \(\int_1^3 (x^2-1)\,dx\) | \(=\) | \(\left[\dfrac{x^3}{3}-x\right]_1^3=\dfrac{20}{3}\) |
| \(A\) | \(=\) | \(\dfrac{2}{3}+\dfrac{20}{3}=\dfrac{22}{3}\) |
The signed integral is \(6\); the total area is \(\dfrac{22}{3}\) square units.
Common pitfalls
Frequently asked questions
What is the difference between a definite integral and the area under a curve?
A definite integral \(\int_a^b f\,dx\) is signed: area below the \(x\)-axis counts as negative. The area between the curve and the axis is \(\int_a^b |f(x)|\,dx\), which is never negative.
How do you integrate the modulus function |f(x)|?
Solve \(f(x)=0\) to find the zeros, split the interval at them, integrate \(f\) on each piece, and add the absolute values of the pieces.
Why is the integral of an odd function over a symmetric interval zero?
For an odd function the graph on \([-a,0]\) is the point reflection of the graph on \([0,a]\), so the negative signed area exactly cancels the positive one, giving \(\int_{-a}^{a} f\,dx=0\).
What happens if you swap the limits of a definite integral?
The sign flips: \(\int_b^a f\,dx=-\int_a^b f\,dx\). Reversing the direction of integration negates the value.
What are the main properties of definite integrals?
Linearity (constants out, sums split), reversed limits flip the sign, \(\int_a^a f\,dx=0\), and splitting over adjacent intervals \(\int_a^b=\int_a^c+\int_c^b\).
Why does the anti-derivative of 1/x use a modulus, ln|x|?
Because \(\dfrac{1}{x}\) is defined for negative \(x\) too, but \(\ln x\) is not. Writing \(\ln|x|\) keeps the argument positive, and differentiating it gives \(\dfrac{1}{x}\) for all \(x\neq0\).