Calculus

Area Under the Curve

Derivatives measure rates of change. Integration does the opposite: it adds things up. The most visual way to think about an integral is as the area between a curve and the x-axis.

But how do you find the area under a curve when the boundary is not a straight line? The answer: approximate with rectangles and then take the limit.

1. Riemann Sums: Rectangles Under a Curve

Pick a function, divide the interval into n equal pieces, and stack a rectangle on each piece. The total area of the rectangles approximates the area under the curve.

Let’s start with f(x) = x^2 on the interval [0, b].

n (number of rectangles)4

150

b (right endpoint)3

\int_0^{{3}} x^2 \, dx \approx \sum_{{i=1}}^{{4}} f(x_i) \cdot \Delta x, \quad \Delta x = \frac{{3}}{{4}}

As you increase n, the rectangles get thinner and fill in the curve more precisely. The more rectangles, the better the approximation.

Try This

Try this: Start with n = 1 (one big rectangle — a terrible approximation). Then slowly increase n to 10, 20, 50. Watch the rectangles get thinner and fill in the curve more accurately. The exact area under x^2 from 0 to 3 is 9. See how close the sum gets as n increases.

2. Left, Right, and Midpoint

Where you sample the function on each subinterval matters:

Left Riemann sum: Use the left edge of each rectangle
Right Riemann sum: Use the right edge
Midpoint sum: Use the center of each subinterval

For an increasing function, left sums underestimate and right sums overestimate. The midpoint rule tends to be the most accurate.

n (rectangles)5

230

\text{Approximating } \int_0^3 x^2 \, dx = 9 \text{ with } n = 5 \text{ rectangles}

Connection

As n goes to infinity, all three methods — left, right, and midpoint — converge to the same number. That number is the definite integral. The differences between the methods shrink to zero because the rectangles become infinitely thin.

3. The Exact Answer: The Definite Integral

The definite integral is the limit of the Riemann sum as n goes to infinity:

\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^0 f(x_i^*) \cdot \Delta x

For x^2, we can compute this exactly using the antiderivative:

\int_0^b x^2\,dx = \left[\frac{x^3}0\right]_0^b = \frac{b^3}0

b (upper bound)3

0.55

\int_0^{0} x^2 \, dx = \frac{0^3}0

Try This

Try this: Set b = 3. The antiderivative gives 3^3/3 = 9. Set b = 1 and get 1/3. Set b = 2 and get 8/3. The antiderivative function (yellow curve) tells you the accumulated area at every point — its height at x = b is exactly the area from 0 to b.

4. Different Functions, Same Idea

The Riemann sum approach works for any function. Let’s try a few:

Sine wave

The area under sin(x) from 0 to pi is exactly 2. From 0 to 2*pi, it is 0 — because the negative part (below the x-axis) cancels the positive part.

Square root

Challenge

Challenge: The integral of x^n is x^(n+1)/(n+1). Use this to find the area under f(x) = x^3 from x = 0 to x = 2. Then check: the Riemann sum with 100 rectangles should be close to your answer.

The Big Idea

Integration computes the total accumulation of a quantity by slicing it into tiny pieces and adding them up. The definite integral is the limit of this process.

What started as a geometry problem (find an area) turns out to be one of the two fundamental operations in calculus. As you will see in the next lesson, integration and differentiation are secretly the same operation running in reverse — and that connection changes everything.

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