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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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