1. Idea of Area Under a Curve
These notes are about how to use integrals to find the area under a graph. When a function \(y = f(x)\) is drawn on the coordinate plane, the region between the curve and the \(x\)-axis often represents an "accumulation" of something: distance, work, charge, or just plain area.
The goal is to turn this region into a number. Instead of trying to count tiny squares, integrals let us add up infinitely many thin rectangles automatically. This is the basic idea behind "area under the curve" using definite integrals.
1.1. From rectangles to integrals
Imagine the graph of \(y = f(x)\) between \(x = a\) and \(x = b\). Divide the interval \([a, b]\) into many small parts of width \(\Delta x\). On each small piece, draw a thin rectangle whose height is approximately \(f(x)\).
The area of one such rectangle is approximately \(f(x) \, \Delta x\). Adding all these small areas gives an approximation of the total area. In the limit, when the number of rectangles goes to infinity and \(\Delta x \to 0\), this sum becomes a definite integral.
\(\text{Area} \approx \sum f(x_i)\, \Delta x \longrightarrow \int_a^b f(x)\, dx\)
2. Definition of Area Under a Curve
Take a real-valued function \(f(x)\) that is continuous on the closed interval \([a, b]\). Suppose that \(f(x) \ge 0\) for all \(x\) in \([a, b]\). Then the area under the curve \(y = f(x)\) from \(x = a\) to \(x = b\), above the \(x\)-axis, is defined as the definite integral
\(\text{Area} = \int_a^b f(x)\, dx.\)
Here, the curve forms the upper boundary, the \(x\)-axis is the lower boundary, and the vertical lines \(x = a\) and \(x = b\) are the left and right boundaries of the region.
2.1. Signed area vs actual area
A definite integral \(\int_a^b f(x)\, dx\) gives the signed area. If \(f(x)\) is positive on an interval, the contribution from that interval is positive. If \(f(x)\) is negative, the contribution is negative.
For pure area problems, the region is usually taken where \(f(x) \ge 0\). If the curve dips below the \(x\)-axis, the integral gives a negative value there, so sometimes one has to split the interval and take absolute values to get actual area. For now, these notes mainly focus on the simple case where \(f(x) \ge 0\) on the interval.
3. Standard Formula for Area Under a Curve
For a continuous function \(f(x)\) on \([a, b]\) with \(f(x) \ge 0\), the standard formula is:
\(\boxed{\text{Area under } y = f(x) \text{ from } x = a \text{ to } x = b = \int_a^b f(x)\, dx}\)
This is the basic formula that appears again and again. Once an antiderivative of \(f(x)\) is known, the area can be calculated using the fundamental theorem of calculus.
3.1. Using the fundamental theorem of calculus
If \(F(x)\) is any antiderivative of \(f(x)\), that is \(F'(x) = f(x)\), then the definite integral is evaluated as
\(\int_a^b f(x)\, dx = F(b) - F(a).\)
This converts the geometric area problem into an algebraic subtraction. The main work is to find \(F(x)\) correctly.
4. Graph Picture and Intuition
It is helpful to always keep a quick sketch of the function in mind. For \(y = f(x)\) on \([a, b]\):
- The \(x\)-axis is drawn horizontally, the \(y\)-axis vertically.
- The curve \(y = f(x)\) is plotted between \(x = a\) and \(x = b\).
- The region of interest is the "curved strip" trapped between the curve and the \(x\)-axis.
- As \(f(x)\) grows taller, the area becomes larger. As the interval \([a, b]\) becomes wider, the area increases as well.
The definite integral acts like a machine that takes this drawn region and outputs a single number that represents its area.
4.1. Thin slices viewpoint
Think of the region under the curve as being made of infinitely many thin vertical slices. Each slice at position \(x\) has height \(f(x)\) and an extremely small width \(dx\). The area of one slice is \(f(x)\, dx\). Adding all slices from \(x = a\) to \(x = b\) gives the integral
\(\int_a^b f(x)\, dx.\)
This thin-slice picture is a good way to remember why integrals appear in area problems.
5. Worked Examples: Area Under a Curve
These examples are meant as reference steps that can be followed whenever an area under a curve problem appears. Each example follows the same pattern: sketch, set up the integral, find an antiderivative, and evaluate.
5.1. Example 1: Area under \(y = x\) from \(0\) to \(2\)
Consider the function \(y = x\) on the interval \([0, 2]\). The graph is a straight line passing through the origin with a positive slope. The region under the curve from \(x = 0\) to \(x = 2\) is a right triangle.
Using integrals, the area is
\(\text{Area} = \int_0^2 x\, dx.\)
An antiderivative of \(x\) is \(\dfrac{x^2}{2}\). So, using the fundamental theorem of calculus:
\(\int_0^2 x\, dx = \left[ \dfrac{x^2}{2} \right]_0^2 = \dfrac{2^2}{2} - \dfrac{0^2}{2} = \dfrac{4}{2} - 0 = 2.\)
So the area under the line \(y = x\) from \(0\) to \(2\) is \(2\) square units. This matches the geometric formula for the area of a triangle with base \(2\) and height \(2\):
\(\text{Area} = \dfrac{1}{2} \times 2 \times 2 = 2.\)
5.2. Example 2: Area under \(y = x^2\) from \(0\) to \(1\)
Now consider the curve \(y = x^2\) on \([0, 1]\). The graph is a parabola opening upwards, touching the origin and lying above the \(x\)-axis in this interval.
To find the area under the curve from \(x = 0\) to \(x = 1\), set up the integral:
\(\text{Area} = \int_0^1 x^2\, dx.\)
An antiderivative of \(x^2\) is \(\dfrac{x^3}{3}\). Applying the limits:
\(\int_0^1 x^2\, dx = \left[ \dfrac{x^3}{3} \right]_0^1 = \dfrac{1^3}{3} - \dfrac{0^3}{3} = \dfrac{1}{3}.\)
So the area under \(y = x^2\) from \(0\) to \(1\) is \(\dfrac{1}{3}\) square units. As a note, this is less than the area under the line \(y = x\) on the same interval, which matches the fact that \(x^2 \le x\) for \(0 \le x \le 1\).
6. Observations and Useful Notes
Some quick points to keep in mind when dealing with area under a curve:
- When the function is non-negative on the interval, the definite integral directly represents the area.
- If the function becomes negative on some part of the interval, the integral gives signed area. For pure area, the region may need to be split and absolute values taken.
- Simpler functions can sometimes be handled using basic geometric shapes (triangles, rectangles, circles), but the integral method is more general and works even when no simple shape is visible.
- The choice of limits \(a\) and \(b\) comes from the description of the region: they are the left and right boundaries along the \(x\)-axis.
- Sketching the graph, even roughly, usually makes it much easier to see which region is being measured and whether the final answer is reasonable.