Preview of the Integral

By the end of this lesson, you will be able to interpret the definite integral as area under a curve, approximate integrals using Riemann sums (left, right, and midpoint), and understand the Fundamental Theorem of Calculus conceptually.

The Big Idea

Just as the derivative answers "what is the instantaneous rate of change?", the integral answers "what is the total accumulation?" Areas, volumes, total distance traveled, work done — all are computed by integration. The integral is the limit of a sum as the number of rectangles approaches infinity.

Area Under a Curve

For a non-negative function $f$ on $[a, b]$, the area between $f$ and the $x$-axis is:

$$A = \int_a^b f(x), dx$$

This integral is the limit of approximating rectangles (Riemann sums) as their width approaches 0.

Riemann Sums

Divide $[a,b]$ into $n$ subintervals of width $\Delta x = \dfrac{b-a}{n}$.

Left Riemann sum: use left endpoint of each subinterval:

$$L_n = \sum_{i=0}^{n-1} f(x_i)\Delta x$$

Right Riemann sum: use right endpoint:

$$R_n = \sum_{i=1}^{n} f(x_i)\Delta x$$

Midpoint sum: use midpoint:

$$M_n = \sum_{i=1}^{n} f\left(\frac{x_{i-1}+x_i}{2}\right)\Delta x$$

As $n \to \infty$ and $\Delta x \to 0$, all three sums converge to the exact area.

Example 1 — Left and Right Riemann Sums

Estimate $\int_0^2 x^2, dx$ using 4 rectangles.

$\Delta x = 0.5$; $x_0 = 0, x_1 = 0.5, x_2 = 1, x_3 = 1.5, x_4 = 2$

Left sum ($L_4$): $f(0) + f(0.5) + f(1) + f(1.5) = 0 + 0.25 + 1 + 2.25 = 3.5$; $L_4 = 3.5 \times 0.5 = 1.75$

Right sum ($R_4$): $f(0.5) + f(1) + f(1.5) + f(2) = 0.25 + 1 + 2.25 + 4 = 7.5$; $R_4 = 7.5 \times 0.5 = 3.75$

True value: $\int_0^2 x^2, dx = [x^3/3]_0^2 = 8/3 \approx 2.67$

Left sum underestimates (function is increasing); right sum overestimates.

Example 2 — Midpoint Sum

$M_4$ for the same integral:

Midpoints: $0.25, 0.75, 1.25, 1.75$

$f(0.25) + f(0.75) + f(1.25) + f(1.75) = 0.0625 + 0.5625 + 1.5625 + 3.0625 = 5.25$

$M_4 = 5.25 \times 0.5 = 2.625$ — much closer to $8/3 \approx 2.667$.