Definite Integrals

Definition

The definite integral of $f$ from $a$ to $b$ is:

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

where $\Delta x = \frac{b-a}{n}$ and $x_i^*$ is a point in the $i$-th subinterval.

Geometric Interpretation

The definite integral represents:

• The signed area between $f(x)$ and the x-axis from $x = a$ to $x = b$
• Area above x-axis is positive
• Area below x-axis is negative

Notation

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

• $a$ = lower limit of integration
• $b$ = upper limit of integration

Properties

Basic Properties

$\int_a^b c \, dx = c(b - a)$

$\int_a^b [f(x) \pm g(x)] \, dx = \int_a^b f(x) \, dx \pm \int_a^b g(x) \, dx$

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

Limits of Integration

$\int_a^a f(x) \, dx = 0$

$\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx$

$\int_a^b f(x) \, dx + \int_b^c f(x) \, dx = \int_a^c f(x) \, dx$

Comparison Properties

If $f(x) \leq g(x)$ on $[a, b]$, then:

$\int_a^b f(x) \, dx \leq \int_a^b g(x) \, dx$

If $m \leq f(x) \leq M$ on $[a, b]$, then:

$m(b - a) \leq \int_a^b f(x) \, dx \leq M(b - a)$

Evaluation Using Antiderivatives

If $F$ is an antiderivative of $f$:

$\int_a^b f(x) \, dx = F(b) - F(a) = \left[F(x)\right]_a^b$

Examples

Example 1: Basic polynomial

$\int_0^2 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3}$

Example 2: Trigonometric

$\int_0^\pi \sin(x) \, dx = \left[-\cos(x)\right]_0^\pi = -\cos(\pi) - (-\cos(0))$

$= -(-1) - (-1) = 1 + 1 = 2$

Example 3: Exponential

$\int_0^1 e^x \, dx = \left[e^x\right]_0^1 = e^1 - e^0 = e - 1$

Example 4: Signed area

$\int_0^{2\pi} \sin(x) \, dx = \left[-\cos(x)\right]_0^{2\pi} = -\cos(2\pi) + \cos(0)$

$= -1 + 1 = 0$

Note: The positive and negative areas cancel out!

Example 5: Using properties

$\int_{-2}^2 (x^3 + 4) \, dx$

Since $x^3$ is odd: $\int_{-2}^2 x^3 \, dx = 0$

$\int_{-2}^2 4 \, dx = 4(2 - (-2)) = 16$

Total: $0 + 16 = 16$

Example 6: Absolute value (total area)

To find the total area between $f(x) = x^2 - 1$ and x-axis on $[-1, 2]$:

$f(x) = 0$ at $x = \pm 1$

$\text{Area} = \left|\int_{-1}^1 (x^2 - 1) \, dx\right| + \int_1^2 (x^2 - 1) \, dx$

$= \left|\left[\frac{x^3}{3} - x\right]_{-1}^1\right| + \left[\frac{x^3}{3} - x\right]_1^2$

$= \left|\left(\frac{1}{3} - 1\right) - \left(-\frac{1}{3} + 1\right)\right| + \left(\frac{8}{3} - 2\right) - \left(\frac{1}{3} - 1\right)$

$= \left|-\frac{4}{3}\right| + \frac{2}{3} + \frac{2}{3}$

$= \frac{4}{3} + \frac{4}{3} = \frac{8}{3}$

Average Value of a Function

The average value of $f$ on $[a, b]$:

$f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx$

Example

Average value of $f(x) = x^2$ on $[0, 3]$:

$f_{\text{avg}} = \frac{1}{3} \int_0^3 x^2 \, dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3}(9) = 3$