Evaluating definite integrals is a fundamental concept under the topic of Integration. This section focuses on methods to evaluate definite integrals, with an emphasis on dealing with improper integrals. Improper integrals occur when the limits of integration are infinite or when the integrand (the function to be integrated) is discontinuous at certain points.

**Understanding Definite Integrals**

A definite integral, denoted as $\int_{a}^{b} f(x) \, dx$, where $f(x)$ is the integrand, and $a$ and $b$ are the limits of integration, calculates the net area under the curve of $f(x)$ from $x = a$ to $x = b$.

**Limits of Integration**: These values, $a$ and $b$, define the interval over which the function is integrated.**Net Area**: The integral calculates the 'net' area, considering areas above the x-axis as positive and below as negative.

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**Techniques for Evaluating Definite Integrals**

Evaluating a definite integral involves finding the antiderivative of the function and then applying the limits of integration.

**1. Find the Antiderivative**

Determine the indefinite integral or antiderivative of $f(x)$.

**2. Apply the Fundamental Theorem of Calculus**

Use the formula $\int_{a}^{b} f(x) \, dx = F(b) - F(a)$, where $F$ is the antiderivative of $f(x)$.

**Example: Evaluating a Basic Definite Integral**

Consider the integral $\int_{0}^{1} x^2 \, dx$:

1. Find the antiderivative of $x^2$, which is $\frac{1}{3}x^3$, using the power rule.

2. Apply the Fundamental Theorem of Calculus:

Evaluate $\frac{1}{3}x^3$ from 0 to 1, resulting in $\frac{1}{3}(1)^3 - \frac{1}{3}(0)^3$.

3. Simplify to get $\frac{1}{3}$.

This represents the area under the curve $x^2$ from $x = 0$ to $x = 1$.

**Improper Integrals**

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Improper integrals arise in two primary scenarios:

**1. Infinite Limits of Integration:**

Either or both limits of a definite integral are infinite (positive or negative).

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**2. Undefined Integrand at a Limit:**

The function to be integrated is undefined at one or both limits of integration.

**Handling Improper Integrals**

To evaluate these integrals:

- Replace the problematic limit with a variable that approaches the problematic value (such as infinity or zero).
- Evaluate the integral with this new limit.
- Finally, consider the behaviour of the integral as the variable approaches its limit.

**Examples**

**Example 1: Undefined Integrand at a Limit**

Evaluate: $\int_0^2 \frac{1}{\sqrt{x}} \, dx$

**Solution:**

Here, the integrand is undefined at $x = 0$. Substitute $0$ with a variable $a$, approaching zero:

$\int_a^2 x^{-\frac{1}{2}} \, dx = \left[\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]_a^2$$= \left[2x^{\frac{1}{2}}\right]_a^2$At $x = 2$, the value is $2 \sqrt{2}$. As $a$ approaches zero, the value approaches $0$.

Thus, the integral is $2\sqrt{2} - 0 = 2\sqrt{2}$.

$\therefore \int_0^2 \frac{1}{\sqrt{x}} \, dx = 2\sqrt{2}$

**Example 2: Infinite Limit of Integration**

Evaluate: $\int_2^{\infty} \frac{1}{4\sqrt{x}} \, dx$

**Solution: **

Here, the upper limit is infinite. Consider an upper limit $b$, approaching infinity:

$\int_2^b \frac{x^{-\frac{1}{2}}}{4} \, dx = \left[\frac{x^{-\frac{1}{2}+1}}{4\left(-\frac{1}{2}+1\right)}\right]_2^b$$= \left[\frac{\sqrt{x}}{2}\right]_2^b$As $b$ approaches infinity, the value of $\frac{\sqrt{x}}{2}$ also approaches infinity.

Thus, the integral is $\infty - \frac{\sqrt{2}}{2} = \infty$.

$\therefore \int_2^{\infty} \frac{1}{4\sqrt{x}} \, dx = \infty$