Area under a curve
Area Under a Curve
Area Under a Curve is the opening topic of Applications of Integration in the California Calculus Standards. It is aligned to Standard 16.0, which requires students to use definite integrals in problems involving area.
The area under a curve from \(a\) to \(b\) is the definite integral \(\int_a^b f(x)\,dx\), the signed area between the curve and the \(x\)-axis; total geometric area uses \(|f|\).
Theory
For a non-negative function, the area under a curve from \(a\) to \(b\) is the definite integral \(\displaystyle\int_a^b f(x)\,dx\). Where the curve dips below the \(x\)-axis the integral counts that area as negative, so total geometric area needs \(|f|\).
The area between \(y=f(x)\ge 0\) and the \(x\)-axis on \([a,b]\) is the definite integral:
Each thin strip has area \(f(x)\,dx\) (height times width); the integral adds them up.
When \(f\) goes below the axis, the integral is a signed area — those pieces subtract. For total geometric area, integrate \(|f|\), splitting at each \(x\)-intercept.
Area under a non-negative curve, and total area for a sign-changing one:
How to find the area under a curve
- Set up \(\displaystyle\int_a^b f(x)\,dx\).
- Antidifferentiate and apply the Fundamental Theorem, \(F(b)-F(a)\).
- For total area, split where \(f=0\) and add the absolute values of the pieces.
Area \(=\displaystyle\int_a^b f\,dx\); antiderivative \(\dfrac{x^3}{3}\), then apply \(F(3)-F(0)\).
| \(\int_0^3 x^2\,dx\) | \(=\) | \(\left[\dfrac{x^3}{3}\right]_0^3\) |
| \(=\) | \(\dfrac{3^3}{3}-\dfrac{0^3}{3}\) | |
| \(=\) | \(9-0=9\) |
Antiderivative \(x^2\); substitute both limits.
| \(\int_0^4 2x\,dx\) | \(=\) | \(\left[x^2\right]_0^4\) |
| \(=\) | \(4^2-0^2\) | |
| \(=\) | \(16\) |
(Check: a triangle of base 4 and height 8 has area \(16\).)
Antiderivative \(\sin x\); use \(\sin\dfrac{\pi}{2}=1\) and \(\sin 0=0\).
| \(\int_0^{\pi/2}\cos x\,dx\) | \(=\) | \(\big[\sin x\big]_0^{\pi/2}\) |
| \(=\) | \(\sin\dfrac{\pi}{2}-\sin 0\) | |
| \(=\) | \(1-0=1\) |
The integral is a signed area, so a piece below the axis subtracts.
For total geometric area, integrate \(|f|\) — split at each zero and add the magnitudes.
Common pitfalls
Frequently asked questions
How do you find the area under a curve?
Integrate the function over the interval: \(\int_a^b f(x)\,dx\), evaluated with an antiderivative via the Fundamental Theorem.
What is signed area?
The definite integral counts area above the \(x\)-axis as positive and area below as negative, so it can be smaller than the geometric area.
How do you find total area when the curve goes below the axis?
Integrate \(|f|\): find the \(x\)-intercepts, integrate over each piece, and add the absolute values.
What are the units of area under a curve?
Square units — the product of the units on the two axes.