# Definite Integrals

## Definite Integrals

Definite Integral is defined as
$\int_{a}^{b} f(x) dx$
• where a is called the lower limit of the integral and b is called the upper limit of the integral.
• Unlike Indefinite integral, it has unique value.
• It can be defined as Integration as Limit of Sum or difference of Antiderivative

## Integration as Limit of Sum

Integration as Limit of Sum is defined as
$\int_{a}^{b} f(x) dx = \lim_{h \rightarrow 0} h[f(a) + f(a+h) + f(a+2h)+.....+f(a + (n-1)h)]$

where $h= \frac {b-a}{n}$
Here The definite integral $\int_{a}^{b} f(x) dx$ is the area bounded by the curve y = f (x), the ordinates x = a, x = b and the x-axis

## Definite Integration as Antiderivative

From the above discussion we can define Area Function as
$A(x)= \int_{a}^{x} f(x) dx$
Here It is the area bounded by the curve y = f (x), the ordinates x = a, x = x and the x-axis
Two Fundamental Theorem of Integral Calculus as given as
Theorem I
$A'(x) = f(x)$

Theorem II
if  $\int f(x) dx= g(x)$
$\int_{a}^{b} f(x) dx =g(b) -g(a)$
This is very important theorem as it makes the Definite Integrals process easiar.You dont need to go through the lengthly process of Integration as Limit of Sum

## Properties of Definite Integrals

$\int_{a}^{b} f(x) dx= \int_{a}^{b} f(t) dt$
$\int_{a}^{b} f(x) dx=- \int_{b}^{a} f(x) dt$
$\int_{a}^{b} f(x) dx= \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$
where a if (x) is a continuous function defined on [0,a],then
$\int_{0}^{a} f(x) dx=\int_{0}^{a} f((a-x)) dx$
$\int_{-a}^{a} f(x) dx=\int_{0}^{a} f((a-x)) dx$
$\int_{-a}^{a} f(x) dx= \begin{cases} 2 \int_{0}^{a} f(x) dx & , f(x) =f(-x) \\ 0 &, f(x) =-f(x) \end{cases}$

## Solved Examples

Example 1
Evaluate $\int_{0}^{1} x^2 \, dx$
Solution
The antiderivative of $x^2$ is:
$\int x^2 \, dx = \frac{x^3}{3} + C$
Now, to find the definite integral, we evaluate this antiderivative at the upper and lower bounds and subtract:
$\int_{0}^{1} x^2 \, dx = \left[\frac{x^3}{3}\right]_0^1$
$= \frac{1^3}{3} - \frac{0^3}{3}$
$= \frac{1}{3}$

Example 2
Evaluate $\int_{-1}^{1} |x| \, dx$
Solution
The function $|x|$ can be broken down as:
$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$
So, the integral becomes:
$\int_{-1}^{1} |x| \, dx = \int_{-1}^{0} (-x) \, dx + \int_{0}^{1} x \, dx$
For the interval $[-1, 0]$:
$\int_{-1}^{0} (-x) \, dx = \left[-\frac{x^2}{2}\right]_{-1}^0 = 0 + \frac{1}{2} = \frac{1}{2}$
For the interval $[0, 1]$:
$\int_{0}^{1} x \, dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2} - 0 = \frac{1}{2}$
Adding them together:
$\int_{-1}^{1} |x| \, dx = \frac{1}{2} + \frac{1}{2} = 1$

Example 3 Evaluate $\int_{0}^{\pi} \sin(x) \, dx$
Solution
The antiderivative of $\sin(x)$ is:
$\int \sin(x) \, dx = -\cos(x) + C$

Evaluating from 0 to $\pi$:
$\int_{0}^{\pi} \sin(x) \, dx = \left[-\cos(x)\right]_0^\pi$
$= -\cos(\pi) + \cos(0)$
$= 1 + 1$
$= 2$

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