The Definite Integral

The definite integral

Introduction to definite integrals

The definite integral of a function represents the signed area under the curve of the function between two specified limits. It calculates the accumulated value (such as area) from a starting point to an endpoint.

The general form of a definite integral is:

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

Where:

• $\int$ is the integral symbol.
• $f(x)$ is the function to integrate.
• $dx$ indicates the variable of integration.
• $a$ is the lower limit of integration.
• $b$ is the upper limit of integration.
• $F(x)$ is the antiderivative of $f(x)$.

Properties of definite integrals

1. Zero width interval:

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

2. Reversing limits:

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

3. Additivity over intervals:

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

4. Constant multiple rule:

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

Step-by-step examples

Example 1: $\int_1^3 2x\, dx$

1. Find the antiderivative of $2x$ using the Power Rule:

   $\int 2x\, dx = x^2 + C$

2. Apply the limits:

   $F(3) - F(1) = (3^2) - (1^2) = 9 - 1$

3. Simplify:

   $8$

4. Final Answer: 8

Example 2: $\int_0^2 x^3\, dx$

1. Find the antiderivative:

   $\int x^3\, dx = \frac{x^4}{4} + C$

2. Apply the limits:

   $F(2) - F(0) = \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} - 0$

3. Simplify:

   $4$

4. Final Answer: 4

Example 3: $\int_0^\pi \sin(x)\, dx$

1. Find the antiderivative:

   $\int \sin(x)\, dx = -\cos(x) + C$

2. Apply the limits:

   $F(\pi) - F(0) = [-\cos(\pi)] - [-\cos(0)]$

3. Evaluate the cosine:

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

4. Simplify:

   $2$

5. Final Answer: 2

Geometric interpretation of definite integrals

The definite integral measures the net area under the curve of a function between two points:

1. Areas above the x-axis are positive.
2. Areas below the x-axis are negative.
3. If necessary, insert a graph showing the shaded region under the curve.

Fundamental theorem of calculus

This theorem connects differentiation and integration:

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

Where $F(x)$ is any antiderivative of $f(x)$.

Example with absolute value function

Example 4: $\int_{-1}^1 |x|\, dx$

1. Split the integral due to the absolute value:

   $\int_{-1}^0 -x\, dx + \int_0^1 x\, dx$

2. Integrate each part:

   $\left(-\frac{x^2}{2}\right) \text{ from } -1 \text{ to } 0 + \left(\frac{x^2}{2}\right) \text{ from } 0 \text{ to } 1$

3. Apply limits:

   $+ [1/2 - 0] = 1/2 + 1/2$

4. Simplify:

   $1$

5. Final Answer: 1

Important notes on definite integrals

1. The result of a definite integral is a number, not a function.
2. The constant of integration $C$ is not included in definite integrals.
3. Areas below the x-axis contribute negatively to the total area.