Finding the exact area: the definite integral
In Year 12 Mathematical Methods (Queensland, QCAA), the fundamental theorem of calculus gives the exact value of a definite integral: if \(F\) is an anti-derivative of \(f\), then \(\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)\). When the curve \(y=f(x)\) is positive on \([a,b]\), this definite integral is the exact area under the curve. You also use the properties of the definite integral — linearity, a zero-width integral, reversing the limits, and splitting the interval.
A definite integral \(\displaystyle\int_a^b f(x)\,dx\) has a lower limit \(a\) and an upper limit \(b\), and produces a single number (not a family of functions). The fundamental theorem of calculus evaluates it: find any anti-derivative \(F\) of \(f\), then subtract its values at the limits, \(F(b)-F(a)\).
Because you subtract, any constant of integration cancels, so a definite integral has no \(+C\). The usual shorthand is \(\big[F(x)\big]_a^b=F(b)-F(a)\).
When \(f(x)>0\) on the interval, the definite integral is the exact area of the region between the curve \(y=f(x)\), the \(x\)-axis and the lines \(x=a\) and \(x=b\), measured in square units. (If the curve dips below the axis the integral gives a signed area, which is a separate topic.)
The fundamental theorem of calculus:
Exact area under a curve that is positive on \([a,b]\):
Standard anti-derivatives used with the theorem:
Properties of the definite integral:
How to evaluate a definite integral / find an exact area
- Find an anti-derivative \(F\) of the integrand \(f\), using the standard forms (no \(+C\) is needed for a definite integral). Write it as \(\big[F(x)\big]_a^b\).
- Substitute the limits. Work out \(F(b)\) (upper limit) and \(F(a)\) (lower limit).
- Subtract in order. The value is \(F(b)-F(a)\) — upper minus lower.
- For an area, first check \(f(x)>0\) on \([a,b]\); then the exact area is \(\displaystyle\int_a^b f(x)\,dx\), stated in square units.
- Use the properties where they help: reverse the limits (change the sign), split the interval, or factor constants out and integrate a sum term by term.
An anti-derivative of \(3x^2\) is \(x^3\).
| \(\displaystyle\int_0^2 3x^2\,dx\) | \(=\) | \(\big[x^3\big]_0^2\) |
| \(=\) | \((2)^3-(0)^3=8\) |
The value of the definite integral is \(8\).
\(y=x^2\ge 0\) on \([0,3]\), so the area is the definite integral. An anti-derivative is \(\dfrac{x^3}{3}\).
| \(A\) | \(=\) | \(\displaystyle\int_0^3 x^2\,dx=\left[\dfrac{x^3}{3}\right]_0^3\) |
| \(=\) | \(\dfrac{27}{3}-0=9\) |
The exact area is \(9\) square units.
An anti-derivative of \(e^{2x}\) is \(\dfrac{1}{2}e^{2x}\) (divide by the coefficient of \(x\)).
| \(\displaystyle\int_0^1 e^{2x}\,dx\) | \(=\) | \(\left[\dfrac{1}{2}e^{2x}\right]_0^1\) |
| \(=\) | \(\dfrac{1}{2}e^2-\dfrac{1}{2}=\dfrac{1}{2}\left(e^2-1\right)\) |
The value is \(\dfrac{1}{2}\left(e^2-1\right)\).
Split \([0,5]=[0,2]\cup[2,5]\); reversing the limits negates the value.
| \(\displaystyle\int_2^5 f\) | \(=\) | \(\displaystyle\int_0^5 f-\int_0^2 f=12-5=7\) |
| \(\displaystyle\int_5^0 f\) | \(=\) | \(-\displaystyle\int_0^5 f=-12\) |
So \(\displaystyle\int_2^5 f=7\) and \(\displaystyle\int_5^0 f=-12\).
Common pitfalls
Frequently asked questions
What is the fundamental theorem of calculus?
If \(F'=f\), then \(\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)\). Find an anti-derivative, substitute the upper then the lower limit, and subtract.
How do you find the exact area under a curve?
If \(y=f(x)>0\) on \([a,b]\), the exact area is \(\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)\), in square units.
Do you add a constant of integration to a definite integral?
No — the constant cancels in \(F(b)-F(a)\). Include \(+C\) only for an indefinite integral.
What happens if you swap the limits?
Reversing the limits negates the value: \(\displaystyle\int_b^a f=-\int_a^b f\). Also \(\displaystyle\int_a^a f=0\).
How do you split a definite integral?
For \(c\) between \(a\) and \(b\), \(\displaystyle\int_a^b f=\int_a^c f+\int_c^b f\), so adjacent areas add.