Chapter 2: Definite Integral

Definite Integral

We study the definition of the definite integral and the Fundamental Theorem of Calculus. We understand its relationship with area.

2.1 The Area Problem

How do we find the area of the region enclosed by the curve $y = f(x)$, the $x$-axis, and the vertical lines $x = a$ and $x = b$?

Idea of Riemann Sum x y y = f(x) a b
Fig.1: Idea of Riemann sum — approximating area under curve with rectangles

Riemann Sum Method

  1. Divide the interval $[a, b]$ into $n$ equal subintervals
  2. Compute the area of the rectangle on each subinterval
  3. Sum the areas of all rectangles
  4. Take the limit as $n \to \infty$

Dividing $[a, b]$ into $n$ equal parts, the partition points are:

$$x_k = a + \frac{b-a}{n}k \quad (k = 0, 1, 2, \ldots, n)$$

The width of each subinterval is $\Delta x = \dfrac{b-a}{n}$

The sum of the rectangle areas over each subinterval $[x_{k-1}, x_k]$ (Riemann sum):

$$S_n = \sum_{k=1}^{n} f(x_k) \cdot \Delta x = \sum_{k=1}^{n} f(x_k) \cdot \frac{b-a}{n}$$

2.2 Definition of the Definite Integral

Definition: Definite Integral

The definite integral of a function $f(x)$ over the interval $[a, b]$ is defined as:

$$\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k) \cdot \frac{b-a}{n}$$

When this limit exists, we say $f(x)$ is integrable on $[a, b]$.

Terminology:

  • $a$: lower limit of integration
  • $b$: upper limit of integration
  • $\displaystyle\int_a^b$: read as "the integral from $a$ to $b$"
Geometric Meaning of the Definite Integral a b
Fig.2: Geometric meaning of definite integral — area = $\displaystyle\int_a^b f(x)\,dx$

Definite vs. Indefinite Integral

  • Indefinite integral $\displaystyle\int f(x)\,dx$: a function (family of antiderivatives)
  • Definite integral $\displaystyle\int_a^b f(x)\,dx$: a number

2.3 Fundamental Theorem of Calculus

Theorem: Fundamental Theorem of Calculus (First Form)

If $f(x)$ is continuous on $[a, b]$ and $F(x)$ is an antiderivative of $f(x)$ (i.e., $F'(x) = f(x)$), then:

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

Notation: We write $F(b) - F(a)$ as $[F(x)]_a^b$ or $\left.F(x)\right|_a^b$.

Thanks to this theorem, computing a definite integral reduces to:

  1. Finding an antiderivative $F(x)$ of $f(x)$
  2. Evaluating $F(b) - F(a)$

There is no need to compute the limit of Riemann sums directly.

Example 2.1

Evaluate $\displaystyle\int_0^2 x^2 \, dx$.

An antiderivative of $x^2$ is $\dfrac{x^3}{3}$.

$$\int_0^2 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} - 0 = \frac{8}{3}$$
x y 0 1 2 y = x²
Fig.3: Definite integral of $y = x^2$ from 0 to 2 (area = $\tfrac{8}{3}$)

Theorem: Fundamental Theorem of Calculus (Second Form)

If $f(x)$ is continuous, then for the function $G(x) = \displaystyle\int_a^x f(t) \, dt$:

$$G'(x) = f(x)$$

This means that differentiating a definite integral with respect to its upper limit recovers the integrand.

2.4 Properties of Definite Integrals

Property 1: Linearity

$$\int_a^b \{cf(x) + dg(x)\} \, dx = c\int_a^b f(x) \, dx + d\int_a^b g(x) \, dx$$

Property 2: Interval Splitting

If $a < c < b$:

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

Property 3: Reversing Limits

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

Property 4: Equal Limits

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

Property 5: Inequality

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

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

Property 6: Even and Odd Functions

If $f(x)$ is an even function ($f(-x) = f(x)$):

$$\int_{-a}^{a} f(x) \, dx = 2\int_0^a f(x) \, dx$$

If $f(x)$ is an odd function ($f(-x) = -f(x)$):

$$\int_{-a}^{a} f(x) \, dx = 0$$
Even function equal equal Odd function neg pos cancel out
Fig.4: Definite integral of even and odd functions on symmetric intervals

2.5 Worked Examples

Example 2.2

Evaluate $\displaystyle\int_1^3 (2x + 1) \, dx$.

$$\begin{align} \int_1^3 (2x + 1) \, dx &= \left[x^2 + x\right]_1^3 \\ &= (3^2 + 3) - (1^2 + 1) \\ &= 12 - 2 = 10 \end{align}$$

Example 2.3

Evaluate $\displaystyle\int_0^{\pi} \sin x \, dx$.

$$\begin{align} \int_0^{\pi} \sin x \, dx &= \left[-\cos x\right]_0^{\pi} \\ &= (-\cos \pi) - (-\cos 0) \\ &= -(-1) - (-1) = 1 + 1 = 2 \end{align}$$

Example 2.4: Using Even Functions

Evaluate $\displaystyle\int_{-2}^{2} x^4 \, dx$.

Since $f(x) = x^4$ is an even function:

$$\int_{-2}^{2} x^4 \, dx = 2\int_0^2 x^4 \, dx = 2\left[\frac{x^5}{5}\right]_0^2 = 2 \cdot \frac{32}{5} = \frac{64}{5}$$

Example 2.5: Using Odd Functions

Evaluate $\displaystyle\int_{-1}^{1} x^3 \, dx$.

Since $f(x) = x^3$ is an odd function:

$$\int_{-1}^{1} x^3 \, dx = 0$$

Verification:

$$\int_{-1}^{1} x^3 \, dx = \left[\frac{x^4}{4}\right]_{-1}^{1} = \frac{1}{4} - \frac{1}{4} = 0 \quad \checkmark$$

Example 2.6: Interval Splitting

Evaluate $\displaystyle\int_0^2 |x - 1| \, dx$.

The sign of $|x - 1|$ changes at $x = 1$:

$$|x - 1| = \begin{cases} 1 - x & (0 \leq x < 1) \\ x - 1 & (1 \leq x \leq 2) \end{cases}$$

Splitting the interval:

$$\begin{align} \int_0^2 |x - 1| \, dx &= \int_0^1 (1 - x) \, dx + \int_1^2 (x - 1) \, dx \\ &= \left[x - \frac{x^2}{2}\right]_0^1 + \left[\frac{x^2}{2} - x\right]_1^2 \\ &= \left(1 - \frac{1}{2}\right) - 0 + \left(2 - 2\right) - \left(\frac{1}{2} - 1\right) \\ &= \frac{1}{2} + 0 + \frac{1}{2} = 1 \end{align}$$

Checklist for Computing Definite Integrals

Check Approach
Does the integrand contain absolute values? Split the interval where the sign changes
Is the integrand even or odd? Use the formula for symmetric intervals
Order of the limits If $a > b$, watch the sign

2.6 Exercises

Problem 1

Evaluate the following definite integrals.

  1. $\displaystyle\int_0^1 (3x^2 - 2x + 1) \, dx$
  2. $\displaystyle\int_1^4 \sqrt{x} \, dx$
  3. $\displaystyle\int_0^{\pi/2} \cos x \, dx$
  4. $\displaystyle\int_1^e \frac{1}{x} \, dx$
Show Solution
  1. $$\left[x^3 - x^2 + x\right]_0^1 = (1 - 1 + 1) - 0 = 1$$
  2. $$\left[\frac{2}{3}x^{3/2}\right]_1^4 = \frac{2}{3}(8 - 1) = \frac{14}{3}$$
  3. $$\left[\sin x\right]_0^{\pi/2} = 1 - 0 = 1$$
  4. $$\left[\ln|x|\right]_1^e = \ln e - \ln 1 = 1 - 0 = 1$$

Problem 2

Evaluate the following definite integrals.

  1. $\displaystyle\int_{-3}^{3} x^5 \, dx$
  2. $\displaystyle\int_{-1}^{1} (x^2 + x^3) \, dx$
  3. $\displaystyle\int_0^3 |2x - 4| \, dx$
Show Solution
  1. Since $x^5$ is an odd function, $\displaystyle\int_{-3}^{3} x^5 \, dx = 0$

  2. $x^2$ is even and $x^3$ is odd, so:

    $$\int_{-1}^{1} (x^2 + x^3) \, dx = \int_{-1}^{1} x^2 \, dx + 0 = 2\int_0^1 x^2 \, dx = 2 \cdot \frac{1}{3} = \frac{2}{3}$$
  3. Since $2x - 4 = 0$ at $x = 2$, the sign changes there:

    $$\begin{align} &= \int_0^2 (4 - 2x) \, dx + \int_2^3 (2x - 4) \, dx \\ &= \left[4x - x^2\right]_0^2 + \left[x^2 - 4x\right]_2^3 \\ &= (8 - 4) - 0 + (9 - 12) - (4 - 8) \\ &= 4 + (-3) - (-4) = 5 \end{align}$$

Problem 3

Evaluate $\displaystyle\frac{d}{dx}\int_0^x t^2 e^t \, dt$.

Show Solution

By the Fundamental Theorem of Calculus (Second Form):

$$\frac{d}{dx}\int_0^x t^2 e^t \, dt = x^2 e^x$$