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:
\[ \int \sin(x) \, dx = -\cos(x) + C \]
